Can You Solve the Olympiad Inequality Challenge with Positive Real Numbers?

[anemone]

Given that $a,\,b$ and $c$ are positive real numbers.

Prove that $$\frac{a^3+b^3+c^3}{3abc}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge 2$$.

 

[anemone]

Hint:

Spoiler

Focus could be put on minimizing $(a+b)(b+c)(c+a)$...

 

[Euge]

Here is my solution.

Spoiler

By the power mean inequality,

$$a^3 + b^3 + c^3 = \frac{a^3 + b^3}{2} + \frac{b^3 + c^3}{2} + \frac{c^3 + a^3}{2} \ge \left(\frac{a + b}{2}\right)^3 + \left(\frac{b + c}{2}\right)^3 + \left(\frac{c + a}{2}\right)^3$$
$$= \frac{(a + b)^3 + (b + c)^3 + (c + a)^3}{8} \ge \frac{3(a + b)(b + c)(c + a)}{8}$$

with equality if and only if $a = b = c$. So

$$\frac{a^3 + b^3 + c^3}{3abc} + \frac{8abc}{(a + b)(b + c)(c + a)} \ge \frac{(a + b)(b + c)(c + a)}{8abc} + \frac{8abc}{(a + b)(b + c)(c + a)} \ge 2$$

using the inequality $x + \frac{1}{x} \ge 2$ with $x = (a + b)(b + c)(c + a)/(8abc)$.

 

[Albert1]

Here is my solution.

Spoiler

By the power mean inequality,

$$a^3 + b^3 + c^3 = \frac{a^3 + b^3}{2} + \frac{b^3 + c^3}{2} + \frac{c^3 + a^3}{2} \ge \left(\frac{a + b}{2}\right)^3 + \left(\frac{b + c}{2}\right)^3 + \left(\frac{c + a}{2}\right)^3$$
$$= \frac{(a + b)^3 + (b + c)^3 + (c + a)^3}{8} \ge \frac{3(a + b)(b + c)(c + a)}{8}$$

with equality if and only if $a = b = c$. So

$$\frac{a^3 + b^3 + c^3}{3abc} + \frac{8abc}{(a + b)(b + c)(c + a)} \ge \frac{(a + b)(b + c)(c + a)}{8abc} + \frac{8abc}{(a + b)(b + c)(c + a)} \ge 2$$

using the inequality $x + \frac{1}{x} \ge 2$ with $x = (a + b)(b + c)(c + a)/(8abc)$.

very innovative!

 

[anemone]

Very well done, Euge!(Cool) And thanks for participating!

My solution:

Spoiler

First note that if we want to minimize the LHS of the target expression, we have to maximize the denominator for $$\frac{8abc}{(a+b)(b+c)(c+a)}$$.

And $$(a+b)(b+c)(c+a)=2abc+a^2b+b^2c+c^2a+a^2c+b^2a+c^2b$$, we have

$$abc\le \frac{a^3+b^3+c^3}{3}$$ by the AM-GM inequality, and both

$$a^2b+b^2c+c^2a\le a^3+b^3+c^3$$, $$a^2c+b^2a+c^2\le a^3+b^3+c^3$$ by the Rearrangement Inequality, thus

$$\begin{align*}\frac{a^3+b^3+c^3}{3abc}+\frac{8abc}{(a+b)(b+c)(c+a)}&= \frac{a^3+b^3+c^3}{3abc}+\frac{8abc}{2abc+a^2b+b^2c+c^2a+a^2c+b^2a+c^2b}\\&\ge \frac{a^3+b^3+c^3}{3abc}+\frac{8abc}{2\left(\frac{a^3+b^3+c^3}{3}\right)+2(a^3+b^3+c^3)}\\&=\frac{a^3+b^3+c^3}{3abc}+\frac{8abc}{\frac{8(a^3+b^3+c^3)}{3}}\\&=\frac{a^3+b^3+c^3}{3abc}+\frac{3abc}{a^3+b^3+c^3}\\&\ge 2\sqrt{\left(\frac{a^3+b^3+c^3}{3abc}\right)\left(\frac{3abc}{a^3+b^3+c^3}\right)}\text{by the AM-GM inequality}\\&=2\end{align*}$$

 

[Euge]

very innovative!

Thank you!