QM-AM-GM-HM Inequality: Formulas, Proofs, and Solved Problems

Suppose $x$ and $y$ are positive real numbers. By using the AM-GM inequality involving $6$ terms, namely $xy,$ $x,$ $y,$ $\dfrac{1}{xy},$ $\dfrac{1}{x},$ and $\dfrac{1}{y},$ the minimum value of the expression is attained when
$$xy = x = y = \dfrac{1}{xy} = \dfrac{1}{x} = \dfrac{1}{y}.$$Since $x, y \in \mathbb{R}^+,$ the only solution to the equation is $x = y = 1.$ This will make the value of $$xy+x+y+\dfrac{1}{xy}+\dfrac{1}{x}+\dfrac{1}{y}$$minimum.
Note:
If you are asked to determine the minimum value, the AM-GM inequality can be used so that we obtain
$$\begin{aligned} \dfrac{xy+x+y+\dfrac{1}{xy}+\dfrac{1}{x}+\dfrac{1}{y}}{6} & \ge \sqrt[6]{xy \cdot x \cdot y \cdot \dfrac{1}{xy} \cdot \dfrac{1}{x} \cdot \dfrac{1}{y}} \\ \dfrac{xy+x+y+\dfrac{1}{xy}+\dfrac{1}{x}+\dfrac{1}{y}}{6} & \ge \sqrt[6]{1} \\ xy+x+y+\dfrac{1}{xy}+\dfrac{1}{x}+\dfrac{1}{y} & \ge 6. \end{aligned}$$Therefore, the minimum value is $\boxed{6}.$
(Answer B)

Suppose $x, y,$ and $z$ are positive real numbers such that $xyz = 1.$ By using the AM-GM inequality, we obtain the following three inequalities.
$$\begin{aligned} x+2y & \ge 2\sqrt{2xy} && (\cdots 1) \\ y+2z & \ge 2\sqrt{2yz} && (\cdots 2) \\ xz + 1 & \ge 2\sqrt{xz} && (\cdots 3) \end{aligned}$$By multiplying the three inequalities side by side, we obtain
$$\begin{aligned} (x+2y)(y+2z)(xz+1) & \ge 2\sqrt{2xy} \cdot 2\sqrt{2yz} \cdot 2\sqrt{xz} \\ & \ge 8\sqrt{4(xyz)^2} \\ & \ge 16(xyz). \end{aligned}$$Since $xyz = 1,$ we obtain $(x+2y)(y+2z)(xz+1) \ge 16.$ This means that the minimum value of $(x+2y)(y+2z)(xz+1)$ is $16.$ This condition is attained when every term involved in Inequalities $1, 2,$ and $3$ has the same value, namely $x = 2y,$ $y = 2z,$ and $xz = 1.$ Simple substitution yields $x = 2,$ $y = 1,$ and $z = \dfrac12.$
(Answer A)

Suppose $a_1, a_2, \cdots, a_{100}$ are positive real numbers and $a_1 + a_2 + \cdots + a_{100} = 2025.$ By using the AM-HM inequality, we obtain the following inequality.
$$\begin{aligned} \dfrac{a_1+a_2+\cdots+a_{100}}{100} & \ge \dfrac{100}{\dfrac{1}{a_1} + \dfrac{1}{a_2} + \cdots + \dfrac{1}{a_{100}}} \\ \dfrac{2.025}{100} & \ge \dfrac{100}{\dfrac{1}{a_1} + \dfrac{1}{a_2} + \cdots + \dfrac{1}{a_{100}}} \\ \dfrac{1}{a_1} + \dfrac{1}{a_2} + \cdots + \dfrac{1}{a_{100}} & \ge \dfrac{10.000}{2.025} = \dfrac{400}{81}. \end{aligned}$$This means that the minimum value of $$\dfrac{1}{a_1} + \dfrac{1}{a_2} + \cdots + \dfrac{1}{a_{100}}$$is $\dfrac{400}{81}.$ This condition is attained when $a_1 = a_2 = \cdots = a_{100} = x$ for some positive real number $x.$ Since $a_1 + a_2 + \cdots + a_{100} = 2.025,$ we obtain
$$\begin{aligned} \underbrace{x + x + \cdots + x}_{\text{100 times}} & = 2025 \\ 100x & = 2025 \\ x & = \dfrac{81}{4}. \end{aligned}$$Therefore, the minimum value of $\dfrac{1}{a_1} + \dfrac{1}{a_2} + \cdots + \dfrac{1}{a_{100}}$ is attained when $a_1 = a_2 = \cdots = a_{100} = \dfrac{81}{4}.$
(Answer C)

Observe that
$$\begin{aligned} \dfrac{4x^2+8x+13}{6+6x} & = \dfrac{4(x^2+2x)+13}{6+6x} \\ & = \dfrac{4((x+1)^2-1)+13}{6+6x} \\ & = \dfrac{4(x+1)^2+9}{6+6x}. \end{aligned}$$Now by using the AM-GM inequality, involving the two terms $\dfrac{4(x+1)^2}{6+6x}$ and $\dfrac{9}{6+6x},$ we obtain
$$\begin{aligned} \dfrac{4(x+1)^2}{6+6x} + \dfrac{9}{6+6x} & \geq 2\sqrt{\dfrac{4(x+1)^2}{6+6x} \cdot \dfrac{9}{6+6x}} \\ \dfrac{4(x+1)^2+9}{6+6x} & \geq 2 \cdot \dfrac{2(x+1)(3)}{6+6x} \\ \dfrac{4(x+1)^2+9}{6+6x} & \geq 2 \cdot \dfrac{\cancel{6x+6}}{\cancel{6+6x}} \\ \dfrac{4(x+1)^2+9}{6+6x} & \geq 2. \end{aligned}$$The last inequality shows that the minimum value of $\dfrac{4x^2+8x+13}{6+6x}$ is $\boxed{2}.$
(Answer C)

Suppose $f(x) = x + \dfrac{1}{x^2} = \dfrac{x}{2} + \dfrac{x}{2} + \dfrac{1}{x^2}$.
By using the AM-GM inequality, involving the $3$ terms $\dfrac{x}{2}, \dfrac{x}{2},$ and $\dfrac{1}{x^2},$ we obtain
$\begin{aligned} \dfrac{x}{2} + \dfrac{x}{2} + \dfrac{1}{x^2} & \geq 3\sqrt[3]{\dfrac{x}{2} \cdot \dfrac{x}{2} \cdot \dfrac{1}{x^2}} \\ x + \dfrac{1}{x^2} & \geq 3\sqrt[3]{\dfrac14}. \end{aligned}$
The last inequality shows that the minimum value of $x + \dfrac{1}{x^2}$ is $\boxed{3\sqrt[3]{\dfrac14}}.$
(Answer B)

It is known that $\color{blue}{xy = \dfrac13}.$
Suppose $x_1 = \dfrac{1}{9x^6}$ and $x_2 = \dfrac{1}{4y^6}$. By using the AM-GM inequality on these two data, we obtain
$\begin{aligned} \textbf{AM} & \geq \textbf{GM} \\ \dfrac{1}{9x^6}+\dfrac{1}{4y^6} & \geq 2\sqrt{\dfrac{1}{9x^6} \cdot \dfrac{1}{4y^6}} \\ & = 2\sqrt{\dfrac{1}{36(xy)^6}} \\ & = 2 \cdot \dfrac{1}{6(\color{blue}{xy})^3} \\ & = \dfrac{1}{3(\frac13)^3} = \dfrac{1}{\frac19} = 9. \end{aligned}$
Therefore, the minimum value of $\dfrac{1}{9x^6}+\dfrac{1}{4y^6}$ is $\boxed{9}.$
(Answer D)

Suppose the two numbers are $x$ and $y$, so that $x+y=24.$ In this case, we will find the minimum value of $\dfrac{1}{x}+\dfrac{1}{y}.$ By using the AM-HM inequality, namely $\textbf{AM} \geq \textbf{HM}$, involving the terms $x$ and $y$, we obtain
$\begin{aligned} \dfrac{x+y}{2} & \geq \dfrac{2}{\frac{1}{x}+\frac{1}{y}} \\ \dfrac{24}{2} & \geq \dfrac{2}{\frac{1}{x}+\frac{1}{y}} \\ 12 & \geq \dfrac{2}{\frac{1}{x}+\frac{1}{y}} \\ \dfrac{1}{x}+\dfrac{1}{y} & \geq \dfrac{2}{12} = \dfrac16. \end{aligned}$
The last inequality shows that the minimum value of the sum of the reciprocals of those numbers is $\boxed{\dfrac16}.$
(Answer E)

It is known that $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=3.$
Suppose $x_1 = \dfrac{a}{b}$, $x_2=\dfrac{b}{c},$ and $x_3=\dfrac{c}{a},$ then based on the AM-GM inequality, we obtain
$\begin{aligned} \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} & \geq 3\sqrt[3]{\dfrac{a}{b} \cdot \dfrac{b}{c} \cdot \dfrac{c}{a}} \\ & = 3\sqrt[3]{1} = 3(1) = 3. \end{aligned}$
However, it is known that $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=3,$ and based on the AM-GM inequality, equality occurs only when $\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{a},$ implying $a = b = c = 1.$
On the other hand, it is known that $a < b < c$ so there cannot exist any value of $a$ that satisfies the condition.
(Answer A)

First simplify the given logarithmic expression using logarithmic properties.
$$\begin{aligned} ^a \log \left(\dfrac{a}{b}\right) + ^b \log \left(\dfrac{b}{a}\right) & = (^a \log a-^a \log b)+(^b \log b-^b \log a) \\ & = (1-^a \log b)+(1-^b \log a) \\ & = 2-(^a \log b +^b \log a) \end{aligned}$$In order for it to attain a maximum value, the value of $^a \log b +^b \log a$ must be made as small as possible. In other words, we must find the minimum value of the expression.
Use the AM-GM inequality.
$\begin{aligned} ^a \log b +^b \log a & \geq 2\sqrt{^a \log b \cdot ^b \log a} \\ & = 2\sqrt{1} = 2 \end{aligned}$
We obtain the minimum value $\color{blue}{2}$. Consequently, the maximum value of $2-(\color{blue}{^a \log b +^b \log a})$ is $2-\color{blue}{2}=0.$
(Answer A)

It is known that
$\begin{aligned} f(x) &= \dfrac{9x^2 \sin^2 x + 4}{x \sin x} \\ & = 9x \sin x + \dfrac{4}{x \sin x}. \end{aligned}$ By using the AM-GM inequality, we obtain
$$\begin{aligned} \textbf{AM} & \geq \textbf{GM} \\ 9x \sin x + \dfrac{4}{x \sin x} & \geq 2\sqrt{(9~\cancel{x \sin x})\left(\dfrac{4}{\cancel{x \sin x}}\right)} \\ & = 2\sqrt{36} = 12. \end{aligned}$$Consequently, $f(x) \geq 12$.
Therefore, the minimum value of $f(x)$ is $\boxed{12}.$
(Answer C)

From $f(x)=x^2+4$, we obtain
$$\begin{aligned} f(xy) & = (xy)^2 + 4 \\ f(y-x) & = (y-x)^2 + 4 = y^2-2xy+x^2+4 \\ f(y+x) & = (y+x)^2+4=y^2+2xy+x^2+4. \end{aligned}$$Substitute each of them into the equation $f(xy)+f(y-x)$ $=f(y+x)$.
$$\begin{aligned} ((xy)^2+4)+(\cancel{y^2}-2xy+\bcancel{x^2+4}) & = \cancel{y^2}+2xy+\bcancel{x^2+4} \\ (xy)^2+4-2xy & = 2xy \\ (xy)^2-4xy+4 & = 0 \\ (xy-2)^2 & = 0 \\ xy & = 2 \end{aligned}$$Next, we will find the minimum value of $x+y$ using the AM-GM inequality and the fact that $\color{blue}{xy=2}$, namely $x+y \geq 2\sqrt{xy} = 2\sqrt{2}.$
Therefore, the minimum value of $x+y$ is $\boxed{2\sqrt2}.$
(Answer E)

It is known that $x^4-2x^3+5x^2-176x+2009=0.$ By using the AM-GM inequality (involving $5$ terms), we obtain
$$\begin{aligned} \dfrac{x^4-2x^3+5x^2-176x+2009}{5} & \geq \sqrt[5]{x^4(-2x^3)(5x^2)(-176x)(2009)} \\ \dfrac{0}{5} & \geq \sqrt[5]{x^{10} \cdot 1760 \cdot 2009} \\ 0 & \geq x^{10} \cdot 1760 \cdot 2009. \end{aligned}$$The inequality above is true when $x^{10}$ is negative or zero. Since $x$ is a real number, $x^{10}$ cannot be negative, meaning the only possibility is that $x^{10}$ must be $0$ so that $x = 0$.
If $x = 0$ is substituted into the polynomial $x^4-2x^3+5x^2-176x+2009$, the result is $0-0+0-0+2009 \neq 0$. Therefore, there does not exist any real number $x$ satisfying the equation $x^4-2x^3+5x^2-176x+2009=0$.
(Answer A)

By using the AM-GM inequality on the $3$ terms: $a^3$, $b^3$, and $1$, we obtain
$\begin{aligned} a^3+b^3+1 & \geq 3\sqrt[3]{a^3(b^3)(1)} \\ a^3+b^3+1 & \geq 3ab. && (\cdots 1) \end{aligned}$
Using the same principle for $b^3$, $c^3$, and $1$, as well as $c^3$, $a^3$, and $1$, we obtain
$\begin{aligned} b^3+c^3+1 & \geq 3bc && (\cdots 2) \\ c^3+a^3+1 & \geq 3ac. && (\cdots 3) \end{aligned}$
Multiply the three equations according to the corresponding sides.
$$\begin{aligned} (a^3+b^3+1)(b^3+c^3+1)(c^3+a^3+1) & \geq (3ab)(3bc)(3ac) \\ (a^3+b^3+1)(b^3+c^3+1)(c^3+a^3+1) & \geq 27a^2b^2c^2 \\ \text{Divide both sides by}~&a^2b^2c^2 \\ \dfrac{(a^3+b^3+1)(b^3+c^3+1)(c^3+a^3+1)}{a^2b^2c^2} & \geq 27 \end{aligned}$$From the last inequality, we obtain that the minimum value of $$\dfrac{(a^3+b^3+1)(b^3+c^3+1)(c^3+a^3+1)}{a^2b^2c^2}$$ is $\boxed{27}.$
(Answer B)

The equation above is equivalent to $a^4+b^4+2=4ab.$ Based on the AM-GM inequality involving the terms $a^4$ and $b^4$, we obtain
$\begin{aligned} a^4+b^4 & \geq 2\sqrt{(a^4)(b^4)} \\ a^4+b^4 & \geq 2(ab)^2 \\ a^4 + b^4 + 2 & \geq 2(ab)^2 + 2. \end{aligned}$
Since $a^4+b^4+2 = 4ab$, we obtain
$\begin{aligned} 2(ab)^2 + 2 & = 4ab \\ 2(ab)^2-4ab+2 & = 0 \\ \text{Divide both sides}&~\text{by}~2 \\ (ab)^2-2ab+1 & = 0 \\ (ab-1)^2 & = 0 \\ ab & = 1. \end{aligned}$
Substitute $ab = 1$ into the original equation to obtain $a^4+b^4 = 4(1)-2 = 2.$ On the other hand, equality occurs when $a^4=b^4$. Thus, we obtain $2b^4 = 2 \Rightarrow b = \pm 1$ and $a = \pm 1$. Observe that $ab = 1$ so that $a$ and $b$ must have the same sign. Therefore, there are $\boxed{2}$ pairs of real numbers $(a, b)$ satisfying the equation, namely $(1, 1)$ and $(-1, -1).$
(Answer C)

Since $a$ and $b$ are positive real numbers, $\dfrac{a}{b}$ is also positive. Let $\dfrac{a}{b} = x,$ then the expression $4\left(\dfrac{a^2}{b^2} + \dfrac{b}{a}\right)^3$ can be written as $4\left(x^2 + \dfrac{1}{x}\right)^3.$
By taking the three terms: $\left(\dfrac{1}{2x^3}, \dfrac{1}{2x^3}, 1\right)$, we apply the AM-GM inequality.
$$\begin{aligned} \dfrac{\dfrac{1}{2x^3}+\dfrac{1}{2x^3}+1}{3} & \geq \sqrt[3]{\dfrac{1}{2x^3} \cdot \dfrac{1}{2x^3} \cdot 1} \\ \dfrac{1}{x^3} + 1 & \geq \dfrac{3}{x^2} \sqrt[3]{\dfrac{1}{4}} \\ \text{Both sides}~&\text{are multiplied by}~x^2 \\ x^2 + \dfrac{1}{x} & \geq \dfrac{3}{\sqrt[3]{4}} \end{aligned}$$Therefore, the minimum value of $x^2 + \dfrac{1}{x}$ is $\color{blue}{\dfrac{3}{\sqrt[3]{4}}}.$
Thus, the minimum value of $4\left(\dfrac{a^2}{b^2} + \dfrac{b}{a}\right)^3$ is given by
$$\boxed{4\left(\dfrac{a^2}{b^2} + \dfrac{b}{a}\right)_{\text{min}}^3 = 4\left(\color{blue}{\dfrac{3}{\sqrt[3]{4}}}\right)^3 = 27}.$$(Answer E)

Since $a, b$ are positive real numbers, we can apply the AM-GM inequality.
First, take the terms $\left(a, \dfrac{4}{a}\right).$
$$\begin{aligned} \textbf{AM} & \geq \textbf{GM} \\ \dfrac{a + \frac{4}{a}}{2} & \geq \sqrt{a \cdot \dfrac{4}{a}} \\ a + \dfrac{4}{a} & \geq 2\sqrt{4} \\ a + \dfrac{4}{a} & \geq 4 && (\cdots 1) \end{aligned}$$Second, take the terms $\left(b, \dfrac{5}{a}\right).$
$$\begin{aligned} \textbf{AM} & \geq \textbf{GM} \\ \dfrac{b + \frac{5}{b}}{2} & \geq \sqrt{b \cdot \dfrac{5}{b}} \\ b + \dfrac{5}{b} & \geq 2\sqrt{5} && (\cdots 2) \end{aligned}$$By multiplying the two inequalities above side by side, we obtain
$$\begin{aligned} \left(a + \dfrac{4}{a}\right)\left(b + \dfrac{5}{b}\right) & \geq 4 \cdot 2\sqrt{5} \\ \left(a + \dfrac{4}{a}\right)\left(b + \dfrac{5}{b}\right) & \geq \sqrt{320}. \end{aligned}$$Equality occurs when the chosen terms in each case are equal.
$$\begin{aligned} a & =\dfrac{4}{a} \Rightarrow a^2 = 4 \Rightarrow a = 2 \\ b & =\dfrac{5}{b} \Rightarrow b^2 = 5 \Rightarrow b = \sqrt{5} \end{aligned}$$Thus, the value of $\boxed{ab = 2\sqrt{5}}.$
(Answer B)

Notice that the expressions $4^{2a}, 4^{2b},$ and $4^{2c}$ are always nonnegative, so we can apply the AM-GM inequality. For the terms $4^{2a}$ and $1,$ we obtain
$$4^{2a} + 1 \ge 2\sqrt{4^{2a}}.$$Similarly, we also obtain
$$\begin{cases} 4^{2b} + 2 \ge 2\sqrt{4^{2a} \cdot 2} \\ 4^{2c} + 8 \ge 2\sqrt{4^{2a} \cdot 8} \end{cases}$$Multiply the three inequalities side by side.
$$\begin{aligned} (4^{2a} + 1)(4^{2b} + 2)(4^{2c} + 8) & \geq (2\sqrt{4^{2a}})(2\sqrt{4^{2a} \cdot 2})(2\sqrt{4^{2a} \cdot 8}) \\ & = 8\sqrt{4^{2a+2b+2c+2}} \\ & = 2^{2a+2b+2c+5} \end{aligned}$$Notice that what we want is for equality to occur, namely $$(4^{2a} + 1)(4^{2b} + 2)(4^{2c} + 8) = 2^{2a + 2b + 2c + 5}.$$This condition occurs when each pair of chosen terms above are equal. From there, we can determine the values of $a, b,$ and $c.$
$$\begin{aligned} 4^{2a} = 1 & \Rightarrow a = 0 \\ 4^{2b} = 2 & \Rightarrow b = 1/4 \\ 4^{2a} = 8 & \Rightarrow c = 3/4 \end{aligned}$$Therefore, the value of $\boxed{\dfrac{a+b}{c} = \dfrac{0 + 1/4}{3/4} = \dfrac13}.$

Take any positive real number $x.$ By using the AM-GM inequality, we obtain the following four inequalities.
$$\begin{aligned} x+1 & \ge 2x^{1/2} \\ x+3 = x+1+1+1 & \ge 4x^{1/4} \\ x+5 = x+1+1+1+1+1 & \ge 6x^{1/6} \\ x+11 = x+\underbrace{1+1+\cdots+1}_{\text{11 terms}} & \ge 12x^{1/12} \end{aligned}$$By multiplying the four inequalities side by side, we obtain
$$\begin{aligned} (x+1)(x+3)(x+5)(x+11) & \ge 2x^{1/2} \cdot 4x^{1/4} \cdot 6x^{1/6} \cdot 12x^{1/12} \\ & = (2 \cdot 4 \cdot 6 \cdot 12) x^{1/2+1/4+1/6+1/12} \\ & = 576x. \end{aligned}$$This means that the greatest value of $M$ is $\boxed{576}.$
(Answer E)

Based on the AM-GM inequality, we obtain
$\begin{aligned} \textbf{AM} & \geq \textbf{GM} \\ \dfrac{\dfrac{x}{y}+\dfrac{y}{x}}{2} & \geq \sqrt{\dfrac{\bcancel{x}}{\cancel{y}} \cdot \dfrac{\cancel{y}}{\bcancel{x}}} = \sqrt1 = 1 \\ \dfrac{x}{y}+\dfrac{y}{x} & \geq 2. \end{aligned}$
Therefore, it is proven that $\dfrac{x}{y} + \dfrac{y}{x} \geq 2$ holds for every $x, y > 0$.

Suppose positive numbers $a, b, c, d$ are given. Based on the AM-HM inequality, we obtain
$$\begin{aligned} \textbf{AM} & \geq \textbf{HM} \\ \dfrac{a+b+c+d}{4} & \geq \dfrac{4}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}} \\ (a+b+c+d)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right) & \geq 4(4) = 16. \end{aligned}$$
Therefore, it is proven that $$(a+b+c+d)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right) \geq 16.$$

Given $\color{red}{p+q+r = 1}$.
We will use the AM-HM inequality, namely $\textbf{AM} \geq \textbf{HM}$.
$\begin{aligned} \dfrac{\color{red}{p+q+r}}{3} & \geq \dfrac{3}{\frac{1}{p}+\frac{1}{q}+\frac{1}{r}} \\ \dfrac{\color{red}{1}}{3} & \geq \dfrac{3}{\frac{1}{p}+\frac{1}{q}+\frac{1}{r}} \\ \dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r} & \geq 3(3) = 9 \end{aligned}$
Therefore, it is proven that $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r} \geq 9.$

Using the AM-GM inequality, the following $3$ statements hold.
$\begin{aligned} \dfrac{a+b}{2} & \geq \sqrt{ab} && (\cdots 1) \\ \dfrac{a+c}{2} & \geq \sqrt{ac} && (\cdots 2) \\ \dfrac{b+c}{2} & \geq \sqrt{bc} && (\cdots 3) \end{aligned}$
Multiply each inequality side by side and we obtain
$$\begin{aligned} \dfrac{a+b}{2} \cdot \dfrac{a+c}{2} \cdot \dfrac{b+c}{2} & \geq \sqrt{ab} \cdot \sqrt{ac} \cdot \sqrt{bc} \\ \dfrac{(a+b)(a+c)(b+c)}{8} & \geq \sqrt{a^2b^2c^2} \\ (a+b)(a+c)(b+c) & \geq 8abc. \end{aligned}$$
Therefore, it is proven that for every $a,b,c \geq 0$, the inequality $(a+b)(a+c)(b+c) \geq 8abc$ holds.

We will use the AM-GM inequality, using the terms $x^2, y^2$, and $z^2$.
For each pair of variables, we obtain
$\begin{aligned} x^2+y^2 & \geq 2\sqrt{x^2 \cdot y^2} = 2xy \\ x^2+z^2 & \geq 2\sqrt{x^2 \cdot z^2} = 2xz \\ y^2+z^2 & \geq 2\sqrt{y^2 \cdot z^2} = 2yz. \end{aligned}$
Add the three inequalities above and we obtain
$\begin{aligned} 2x^2+2y^2+2z^2 & \geq 2xy + 2xz + 2yz \\ \text{Divide both sides}&~\text{by}~2 \\ x^2+y^2+z^2 & \geq xy+xz+yz. \end{aligned}$
The statement is proven.
Equality occurs when $x^2 = y^2 = z^2$, namely when $x = y = z = 1$.

Notice that
$999! = 1 \times 2 \times 3 \times \cdots \times 999$.
Based on the AM-GM inequality, we have
$$\boxed{\sqrt[n]{a_1 \times a_2 \times a_3 \times \cdots \times a_{n-1} \times a_n} \leq \dfrac{\sum_{i=1}^n a_n}{n}}.$$Equality holds if and only if $a_1=a_2=a_3=\cdots=a_{n-1}=a_n.$
Since $a_1 \neq a_2 \neq a_3 \neq \cdots \neq a_{n-1} \neq a_n$, we obtain
$$\begin{aligned} \sqrt[n]{a_1 \times a_2 \times a_3 \times \cdots \times a_{n-1} \times a_n} & < \dfrac{\sum_{i=1}^n a_n}{n} \\ \sqrt[999]{1 \times 2 \times 3 \times \cdots \times 999} & < \dfrac{\color{blue}{1+2+3+\cdots+999}}{999}. \end{aligned}$$The series marked in blue above is an arithmetic series. Its sum can be found using the formula $\text{S}_n = \dfrac{n}{2}(a+\text{U}_n)$.
We obtain the inequality
$\begin{aligned} \sqrt[999]{999!} & < \dfrac{\dfrac{999}{2}(1+999)}{999} \\ \sqrt[999]{999!} & < \dfrac{\cancel{999} \cdot 500}{\cancel{999}} \\ \sqrt[999]{999!} & < 500 \\ 999! & < 500^{999}. \end{aligned}$
It is proven that $999! < 500^{999}$.

Using the AM-GM inequality, taking the terms $a, \underbrace{b, b, \cdots, b}_{n~\text{times}}$, we obtain
$\begin{aligned} \dfrac{a+\overbrace{b+b+\cdots+b}^{n~\text{times}}}{n+1} & \geq (a \cdot \overbrace{b \cdot b \cdots b}^{n~\text{times}})^{\frac{1}{n+1}} \\ \dfrac{a+nb}{n+1} & \geq (ab^n)^{\frac{1}{n+1}} \\ \left(\dfrac{a+nb}{n+1}\right)^{n+1} & \geq ab^n. \end{aligned}$
Therefore, the inequality is proven to be true.

Observe that
$$\begin{aligned} \dfrac{a+1}{a}+\dfrac{b+1}{b} & = \dfrac{a+(a+b)}{a}+\dfrac{b+(a+b)}{b} \\ & = 4 + \dfrac{a}{b} + \dfrac{b}{a}. \end{aligned}$$Using the AM-GM inequality, taking the two terms $\dfrac{a+1}{a}$ and $\dfrac{b+1}{b}$, we obtain
$$\begin{aligned} \dfrac{a+1}{a} + \dfrac{b+1}{b} = 4 + \dfrac{a}{b}+\dfrac{b}{a} & \geq 4 + 2\sqrt{\dfrac{a}{b} \cdot \dfrac{b}{a}} \\ 4 + \dfrac{a}{b}+\dfrac{b}{a} & \geq 4+2 = 6. \end{aligned}$$Next, using the QM-AM inequality, taking the two terms $\dfrac{a+1}{a}$ and $\dfrac{b+1}{b}$, we obtain
$$\begin{aligned} \left(\dfrac{\left(\dfrac{a+1}{a}\right)^2+\left(\dfrac{b+1}{b}\right)^2}{2}\right)^{1/2} & \geq \dfrac{\dfrac{a+1}{a}+\dfrac{b+1}{b}}{2} \\ \left(\dfrac{\left(\dfrac{a+1}{a}\right)^2+\left(\dfrac{b+1}{b}\right)^2}{2}\right)^{1/2} & \geq \dfrac{6}{2} = 3 \\ \dfrac{\left(\dfrac{a+1}{a}\right)^2+\left(\dfrac{b+1}{b}\right)^2}{2} & \geq 9 \\ \left(\dfrac{a+1}{a}\right)^2+\left(\dfrac{b+1}{b}\right)^2 & \geq 18. \end{aligned}$$Therefore, it is proven that $\left(\dfrac{a+1}{a}\right)^2+\left(\dfrac{b+1}{b}\right)^2 \geq 18$.

Given that $\color{blue}{a+b=1}$.
Suppose $f(a, b)$ is a function of two variables.
$\begin{aligned} f(a, b) & = \left(\dfrac{a^2+1}{a}\right)^2+\left(\dfrac{b^2+1}{b}\right)^2 \\ & = \left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2 \\ & = a^2+2+\dfrac{1}{a^2} + b^2 + 2 + \dfrac{1}{b^2} \\ & = a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+4 \end{aligned}$
We will find the minimum value of $f(a, b)$, which is equivalent to finding the minimum value of $g(a, b) = a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}$.
Using the QM-AM inequality, taking the $2$ terms $a$ and $b$, we obtain
$\begin{aligned} \sqrt{\dfrac{a^2 + b^2}{2}} & \geq \dfrac{a+b}{2} \\ \text{Square both sides} \\ \dfrac{a^2+b^2}{2} & \geq \left(\dfrac{\color{blue}{a+b}}{2}\right)^2 \\ \dfrac{a^2+b^2}{2} & \geq \left(\dfrac{1}{2}\right)^2 = \dfrac14 \\ \color{red}{a^2+b^2} & \color{red}{\geq \dfrac12}. \end{aligned}$
If we use the GM-HM inequality with the terms $a^2$ and $b^2$, we obtain
$\sqrt{a^2b^2} = ab \geq \dfrac{2}{\dfrac{1}{a^2}+\dfrac{1}{b^2}}.~~~~(\cdots 1)$
If we use the AM-GM inequality with the terms $a$ and $b$, we obtain
$\begin{aligned} \dfrac{\color{blue}{a+b}}{2} & \geq \sqrt{ab} \\ \dfrac12 & \geq \sqrt{ab} \\ \text{Square both sides} \\ \dfrac14 & \geq ab. && (\cdots 2) \end{aligned}$
From inequalities $(1)$ and $(2)$, we obtain
$\begin{aligned} \dfrac14 & \geq \dfrac{2}{\dfrac{1}{a^2}+\dfrac{1}{b^2}} \\ \color{red}{\dfrac{1}{a^2}+\dfrac{1}{b^2}} & \color{red}{\geq 2(4) = 8}. \end{aligned}$
Now add the $2$ inequalities marked in red.
$a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2} \geq 8+\dfrac12 = \dfrac{17}{2}$
The minimum value of $g(a, b) = a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}$ is $\dfrac{17}{2}$, which means the minimum value of $g(a,b) = f(a,b)+4$ is $\dfrac{17}{2}+4$ $=\dfrac{25}{2}$.
Therefore, it is proven that $\left(\dfrac{a^2+1}{a}\right)^2+\left(\dfrac{b^2+1}{b}\right)^2 \geq \dfrac{25}{2}.$

Consider the form
$\begin{aligned} \dfrac{1}{a} + \dfrac{1}{a+4} & = \dfrac{(a+4)+a}{a(a+4)} \\ & = \dfrac{2a+4}{a(a+4)} \\ & = 2 \cdot \dfrac{a+2}{a(a+4)} \\ \Leftrightarrow \dfrac12\left(\dfrac{1}{a} + \dfrac{1}{a+4}\right) & = \dfrac{a+2}{a(a+4)}. \end{aligned}$
Therefore,
$$\begin{aligned} & \dfrac{a+2}{a(a+4)}+\dfrac{b+2}{b(b+4)}+\dfrac{c+2}{c(c+4)} \\ & = \dfrac12\left(\dfrac{1}{a}+\dfrac{1}{a+4} + \dfrac{1}{b}+\dfrac{1}{b+4} + \dfrac{1}{c}+\dfrac{1}{c+4}\right) \\ & = \dfrac12\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a+4}+\dfrac{1}{b+4}+\dfrac{1}{c+4}\right). \end{aligned}$$Using the AM-HM inequality with the terms $a, b$, and $c$, we obtain
$\begin{aligned} \dfrac{a+b+c}{3} & \geq \dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} \\ \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} & \geq \dfrac{9}{a+b+c}. \end{aligned}$
Using the AM-HM inequality with the terms $a+4, b+4$, and $c+4$, we obtain
$$\begin{aligned} \dfrac{a+4+b+4+c+4}{3} & \geq \dfrac{3}{\dfrac{1}{a+4}+\dfrac{1}{b+4}+\dfrac{1}{c+4}} \\ \dfrac{a+b+c+12}{3} & \geq \dfrac{3}{\dfrac{1}{a+4}+\dfrac{1}{b+4}+\dfrac{1}{c+4}} \\ \dfrac{1}{a+4}+\dfrac{1}{b+4}+\dfrac{1}{c+4} & \geq \dfrac{9}{a+b+c+12}. \end{aligned}$$Now we obtain
$$\begin{aligned} & \dfrac{a+2}{a(a+4)}+\dfrac{b+2}{b(b+4)}+\dfrac{c+2}{c(c+4)} \\ & = \dfrac12\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a+4}+\dfrac{1}{b+4}+\dfrac{1}{c+4}\right) \\ & \geq \dfrac12\left(\dfrac{9}{a+b+c}+\dfrac{9}{a+b+c+12}\right) \\ & \geq \dfrac12\left(\dfrac{9}{6}+\dfrac{9}{6+12}\right) \\ & = \dfrac12\left(\dfrac32 + \dfrac12\right) = 1. \end{aligned}$$Therefore, it is proven that
$$\dfrac{a+2}{a(a+4)}+\dfrac{b+2}{b(b+4)}+\dfrac{c+2}{c(c+4)} \geq 1.$$

Based on the AM-GM inequality using the terms $1$ and $x$, we obtain
$$\begin{aligned} 1+x & \geq 2\sqrt{(1)(x)} = 2\sqrt{x} \\ 1+x & \geq (1+\sqrt{x})^2-(1+x) \\ (1+x)+(1+x) & \geq (1+\sqrt{x})^2 \\ 2(1+x) & \geq (1+\sqrt{x})^2 \\ \dfrac{1}{(1+\sqrt{x})^2} & \geq \dfrac{1}{2(1+x)}. && (\cdots 1)\end{aligned}$$Using the same principle, but for the terms $y$ and $1$, we obtain
$\dfrac{1}{(1+\sqrt{y})^2} \geq \dfrac{1}{2(1+y)}.~~~~(\cdots 2)$
Adding the two inequalities above, we get
$$\begin{aligned} \dfrac{1}{(1+\sqrt{x})^2} + \dfrac{1}{(1+\sqrt{y})^2} & \geq \dfrac{1}{2(1+x)}+\dfrac{1}{2(1+y)} \\ & = \dfrac12\left(\dfrac{1}{1+x}+\dfrac{1}{1+y}\right). && (\cdots 3) \end{aligned}$$Next, use the AM-HM inequality with the terms $\dfrac{1}{1+x}$ and $\dfrac{1}{1+y}$.
$$\begin{aligned} \dfrac{1}{1+x}+\dfrac{1}{1+y} & \geq 2 \cdot \dfrac{2}{\dfrac{1}{\frac{1}{1+x}} + \dfrac{1}{\frac{1}{1+y}}} \\ \dfrac{1}{1+x}+\dfrac{1}{1+y} & \geq \dfrac{4}{1+x+1+y} \\ \dfrac{1}{1+x}+\dfrac{1}{1+y} & \geq \dfrac{4}{x+y+2} && (\cdots 4) \end{aligned}$$We obtain the relationship between inequalities $(3)$ and $(4)$, namely
$$\begin{aligned} \dfrac{1}{(1+\sqrt{x})^2} + \dfrac{1}{(1+\sqrt{y})^2} & \geq \dfrac12\left(\dfrac{1}{1+x}+\dfrac{1}{1+y}\right) \\ \Rightarrow \dfrac{1}{(1+\sqrt{x})^2} + \dfrac{1}{(1+\sqrt{y})^2} & \geq \dfrac12\left(\dfrac{4}{x+y+2}\right) = \dfrac{2}{x+y+2}. \end{aligned}$$Therefore, it is proven that $$\dfrac{1}{(1+\sqrt{x})^2} + \dfrac{1}{(1+\sqrt{y})^2} \geq \dfrac{2}{x+y+2}.$$

Claim that $\sqrt{(a+b)(a+c)} \geq \sqrt{ab} + \sqrt{ac}.$
In fact, squaring both sides yields the inequality
$$\begin{aligned} \left(\sqrt{(a+b)(a+c)}\right)^2 & \geq \left(\sqrt{ab} + \sqrt{ac}\right)^2 \\ (a+b)(a+c) & \geq ab+2a\sqrt{bc} + ac \\ a^2+\cancel{ac+ab}+bc & \geq \cancel{ab}+2a\sqrt{bc} + \cancel{ac} \\ a^2+bc & \geq 2a\sqrt{bc}. \end{aligned}$$The last inequality follows from the AM-GM relation using the terms $a^2$ and $bc$. Therefore, the previous claim is true.
Since $\sqrt{(a+b)(a+c)} \geq \sqrt{ab} + \sqrt{ac}$, we obtain
$$\begin{aligned} \dfrac{a}{a + \sqrt{(a+b)(a+c)}} & \leq \dfrac{a}{a+\sqrt{ab}+\sqrt{ac}} \\ & = \dfrac{a}{\sqrt{a}(\sqrt{a}+\sqrt{b}+\sqrt{c}} \\ & = \dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}. && (\cdots 1) \end{aligned}$$Similarly, we obtain
$$\begin{aligned} \dfrac{b}{b + \sqrt{(b+c)(b+a)}} & \leq \dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}} && (\cdots 2) \\ \dfrac{c}{c + \sqrt{(c+a)(c+b)}} & \leq \dfrac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}. && (\cdots 3) \end{aligned}$$Add inequalities $(1)$, $(2)$, and $(3)$ so that we obtain
$$\begin{aligned} & \dfrac{a}{a + \sqrt{(a+b)(a+c)}} + \dfrac{b}{b + \sqrt{(b+c)(b+a)}} + \dfrac{c}{c + \sqrt{(c+a)(c+b)}} \\ & \leq \dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}} + \dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}} + \dfrac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}} \\ & = \dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}} = 1. \end{aligned}$$Therefore, the given inequality has been proven.

Given $a + b = ab$.
Squaring both sides, we obtain
$\begin{aligned} (a + b)^2 & = (ab)^2 \\ a^2+b^2+2ab & = (ab)^2. \end{aligned}$
According to the AM-GM inequality, we have $a^2+b^2 \geq 2ab$ so that we further obtain
$\begin{aligned} a^2+b^2+\color{red}{+2ab} & \geq 2ab\color{red}{+2ab} \\ (ab)^2 & \geq 4ab \\ ab & \geq 4. \end{aligned}$
Therefore, we get
$$\begin{aligned} \dfrac{a}{b^2+2017}+\dfrac{b}{a^2+2017} & = \dfrac{a^2}{ab^2+2017a} + \dfrac{b^2}{a^2b + 2017b} \\ & \geq \dfrac{(a+b)^2}{ab(a+b)+2017(a+b)} \\ & = \dfrac{(ab)^2}{ab^2+2017ab} \\ & = \dfrac{ab}{ab+2017} = 1-\dfrac{2017}{ab+2017} \\ & \geq 1-\dfrac{2017}{4+2017} = \dfrac{4}{2021}. \end{aligned}$$Therefore, it is proven that $\dfrac{a}{b^2+2017}+\dfrac{b}{a^2+2017} \geq \dfrac{4}{2021}$.

From the AM-GM inequality that $ab \leq \left(\dfrac{a+b}{2}\right)^2$ for positive integers $a,b$, we obtain the following inequalities.
$$\begin{aligned} 1 \cdot n & \leq \left(\dfrac{1+n}{2}\right)^2 && (\text{if}~a = 1, b = n) \\ 2 \cdot (n-1) & \leq \left(\dfrac{1+n}{2}\right)^2 && (\text{if}~a = 2, b = n-1) \\ 3 \cdot (n-2) & \leq \left(\dfrac{1+n}{2}\right)^2 && (\text{if}~a = 3, b = n-2) \\ \hspace{8pt} & \vdots && \hspace{5.5pt} \vdots \\ (n-1) \cdot 2 & \leq \left(\dfrac{1+n}{2}\right)^2 && (\text{if}~a = n-1, b = 2) \\ n \cdot 1 & \leq \left(\dfrac{1+n}{2}\right)^2 && (\text{if}~a = n, b = 1) \end{aligned}$$If all the forms above are multiplied according to their corresponding sides so that the inequality sign is preserved, we obtain
$$\begin{aligned} (1 \cdot n)(2 \cdot (n-1))(3 \cdot (n-2)) \cdots ((n-1) \cdot 2)(n \cdot 1) & \leq \underbrace{\left(\dfrac{1+n}{2}\right)^2\left(\dfrac{1+n}{2}\right)^2\cdots\left(\dfrac{1+n}{2}\right)^2}_{\text{as many as}~n} \\ (1 \cdot 2 \cdot 3 \cdots n)(n(n-1)(n-2)\cdots 1) & \leq \left(\dfrac{1+n}{2}\right)^{2n} \\ n! \cdot n! & \leq \left(\dfrac{1+n}{2}\right)^{2n} \\ (n!)^2 & \leq \left(\dfrac{1+n}{2}\right)^{2n} \\ n! & \leq \left(\dfrac{1+n}{2}\right)^n. \end{aligned}$$Therefore, it is proven that $\bbox[skyblue, 3pt]{n! \leq \left(\dfrac{n+1}{2}\right)^n}.$

Given $a, b > 0$ and $\color{red}{ab = 6}.$
Take the three terms $\left(2a^2, 2b, 2b\right).$ Using the AM-GM inequality, we obtain
$$\begin{aligned} \textbf{AM} & \geq \textbf{GM} \\ \dfrac{2a^2 + 2b + 2b}{3} & \geq \sqrt[3]{2a^2 \cdot 2b \cdot 2b} \\ 2a^2 + 4b & \geq 3 \sqrt[3]{8a^2b^2} \\ 2a^2 + 4b & \geq 6\sqrt[3]{(\color{red}{ab})^2} \\ 2a^2 + 4b & \geq 6\sqrt[3]{36}. \end{aligned}$$The last inequality shows that the maximum value of $2a^2 + 4b$ is $\boxed{6\sqrt[3]{36}}.$
Equality occurs when the chosen terms are equal, namely $2a^2 = 2b$ or simplified to $a^2 = b.$ Since $ab = 6,$ substitution gives
$$\begin{aligned} a(a^2) & = 6 \\ a^3 & = 6 \\ a & = \sqrt[3]{6} \\ \Rightarrow b & = \sqrt[3]{36}. \end{aligned}$$Therefore, the values of $a$ and $b$ for which the minimum value is attained are respectively $\boxed{\sqrt[3]{6}}$ and $\boxed{\sqrt[3]{36}}.$

Given $a, b > 0$ and $a + 4b = 12.$
Answer a)
Take the two terms $\left(a, 4b\right).$ Using the AM-GM inequality, we obtain
$$\begin{aligned} \textbf{AM} & \geq \textbf{GM} \\ \dfrac{a + 4b}{2} & \geq \sqrt{a \cdot 4b} \\ \dfrac{12}{2} & \geq \sqrt{4ab} \\ 6 & \geq \sqrt{4ab} \\ 36 & \geq 4ab \\ 9 & \geq ab. \end{aligned}$$The last inequality shows that the maximum value of $ab$ is $\boxed{9}$ (attained when $a = 4b).$
Answer b)
Take the three terms $\left(a, 2b, 2b\right).$ Using the AM-GM inequality, we obtain
$$\begin{aligned} \textbf{AM} & \geq \textbf{GM} \\ \dfrac{a + 2b + 2b}{3} & \geq \sqrt[3]{a \cdot 2b \cdot 2b} \\ \dfrac{12}{3} & \geq \sqrt[3]{4ab^2} \\ 4 & \geq \sqrt[3]{4ab^2} \\ 64 & \geq 4ab^2 \\ 16 & \geq ab^2. \end{aligned}$$The last inequality shows that the maximum value of $ab^2$ is $\boxed{16}$ (attained when $a = 2b).$
Note:
If the three chosen terms are $\left(a, b, 3b\right),$ then by using the same method as above, we obtain
$$\begin{aligned} \textbf{AM} & \geq \textbf{GM} \\ \dfrac{a + b + 3b}{3} & \geq \sqrt{a \cdot b \cdot 3b} \\ \dfrac{12}{3} & \geq \sqrt[3]{3ab^2} \\ 4 & \geq \sqrt[3]{3ab^2} \\ 64 & \geq 3ab^2 \\ \dfrac{64}{3} & \geq ab^2. \end{aligned}$$Observe that equality occurs when all chosen terms are equal, namely $a = b = 3b,$ but this condition is impossible for positive real numbers $a, b.$ Therefore, we cannot conclude that the maximum value of $ab^2$ is $\dfrac{64}{3}.$

First, take the three terms $\left(a^3, a^3, b^3\right).$
Using the AM-GM inequality, we obtain
$$\begin{aligned} \textbf{AM} & \geq \textbf{GM} \\ \dfrac{a^3 + a^3 + b^3}{3} & \geq \sqrt[3]{a^3 \cdot a^3 \cdot b^3} \\ 2a^3 + b^3 & \geq 3\sqrt[3]{a^6b^3} \\ 2a^3 + b^3 & \geq 3a^2b. && (\cdots 1) \end{aligned}$$Second, take the three terms $\left(b^3, b^3, c^3\right).$
Using the AM-GM inequality, we obtain
$$\begin{aligned} \textbf{AM} & \geq \textbf{GM} \\ \dfrac{b^3 + b^3 + c^3}{3} & \geq \sqrt[3]{b^3 \cdot b^3 \cdot c^3} \\ 2b^3 + c^3 & \geq 3\sqrt[3]{b^6c^3} \\ 2b^3 + c^3 & \geq 3b^2c. && (\cdots 2) \end{aligned}$$Finally, take the three terms $\left(c^3, c^3, a^3\right).$
Using the AM-GM inequality, we obtain
$$\begin{aligned} \textbf{AM} & \geq \textbf{GM} \\ \dfrac{c^3 + c^3 + a^3}{3} & \geq \sqrt[3]{c^3 \cdot c^3 \cdot a^3} \\ 2c^3 +a^3 & \geq 3\sqrt[3]{c^6a^3} \\ 2c^3 + a^3 & \geq 3c^2a. && (\cdots 3) \end{aligned}$$Add the three inequalities according to their corresponding sides to obtain
$$\begin{aligned} (2a^3 + b^3) + (2b^3 + c^3) + (2c^3 + a^3) & \geq 3a^2b + 3b^2c + 3c^2a \\ 3a^3 + 3b^3 + 3c^3 & \geq 3a^2b + 3b^2c + 3c^2a \\ a^3 + b^3 + c^3 & \geq a^2b + b^2c + c^2a. \end{aligned}$$Therefore, it is proven that for positive real numbers $a, b, c,$ we have $$a^3 + b^3 + c^3 \geq a^2b + b^2c + c^2a.$$