Let $a,b,c > 0 : abc=1$. Prove that $24\sum\limits_{cyc}\dfrac{a^5}{b}+7\sum\limits_{cyc}\dfrac{b}{a^5} \ge 31(a^5+b^5+c^5).$

Question

My friend sent me a excessive hard inequality problem, it's

Let $a,b,c > 0 : abc=1.$ Prove that $$24\left(\dfrac{a^5}{b}+\dfrac{b^5}{c}+\dfrac{c^5}{a}\right)+7\left(\dfrac{b}{a^5}+\dfrac{c}{b^5}+\dfrac{a}{c^5}\right) \ge 31(a^5+b^5+c^5).$$

I posted it on AOPS: here Besides, I also checked the correctness of this problem by software.

Here is my attempts Firstly, Let $a=\dfrac{x}{y}, b=\dfrac{y}{z}, c=\dfrac{z}{x},$ the inequality equivalent to $$24\sum\limits_{cyc}\dfrac{x^5z}{y^6}+7\sum\limits_{cyc}\dfrac{y^6}{x^5z} \ge 31\sum\limits_{cyc}\dfrac{x^5}{y^5}.$$ Subsequently, by AM-GM inequality, we obtain $$24\dfrac{x^5z}{y^6}+7\dfrac{y^6}{x^5z} \ge 31\sqrt[31]{\dfrac{x^{120}z^{17}}{y^{137}}},$$ Hence, we need to prove that $$\sum\limits_{cyc}\sqrt[31]{\dfrac{x^{120}z^{17}}{y^{137}}} \ge \sum\limits_{cyc}\dfrac{x^5}{y^5}.$$ And I am stuck here, so I need more idea to prove this.

Answer 1

Hint: Applying the AM-GM inequality to

• $24$ times the term $\dfrac{a^5}{b}$,
• $1$ times the term $\dfrac{c}{b^5}$,
• $6$ times the term $\dfrac{a}{c^5}$

we have then:

$$\begin{align}24\cdot \dfrac{a^5}{b}+\dfrac{c}{b^5}+6\cdot \dfrac{a}{c^5} \ge31\sqrt[31]{\left(\dfrac{a^5}{b}\right)^{24}\cdot\left(\dfrac{c}{b^5}\right)\cdot\left(\dfrac{a}{c^5}\right)^6}&=31\sqrt[31]{a^{126}b^{-29}c^{-29}} \\&= 31\sqrt[31]{a^{155}(abc)^{-29}}\\&=31a^5 \end{align}$$

The equality occurs if and only if $a = b=c = 1$.