# Proving $\sum_{cyc} \frac{1}{a^6 + b^6 + 3c^3 + 4} \leq \frac{3}{3 + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca})}$

Problem proposed for JBMO practice in symmetrical inequalities (Chebyshev, rearrangement):

For every positive real numbers $$a, b, c$$, for which $$a + b + c = 3$$ we have: $$\sum_{cyc} \frac{1}{a^6 + b^6 + 3c^3 + 4} \leq \frac{3}{3 + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca})}$$

My attempt:

Step 1
AM-GM states that $$a^6 + b^6 \geq 2a^3b^3$$
Therefore, $$LHS \leq \sum_{cyc} \frac{1}{2a^3b^3 + 2c^3 + 2 + c^3 + 2}$$

Step 2
AM-GM states that $$2a^3b^3 + 2c^3 + 2 \geq 6abc$$ and that $$c^3 + 1 + 1 \geq 3c$$
This means that $$LHS \leq \sum_{cyc} \frac{1}{6abc + 3a} = \frac{1}{3} \sum_{cyc} \frac{1}{2abc + a}$$

Step 3
Rearrangement inequality states that $$\sqrt{ab} + \sqrt{bc} + \sqrt{ca} \leq a + b + c = 3$$ This means: $$RHS \geq \frac{1}{3}$$

Remaining of the proof and further ideas
We should prove $$\sum_{cyc} \frac{1}{2abc + a} \leq 1$$
I should probably apply Chebyshev but I could get no good result after applying its variants...

• the last inequality is false (what happens when $\lim_{a\to 0}$) – Albus Dumbledore Dec 9 '20 at 14:12
• Exactly. The last inequality should be in fact $\displaystyle \sum_{cyc} \dfrac{1}{2abc+a} \geq 1$ (Take for example, $a=b=\frac12,c=2)$. – Light Yagami Dec 9 '20 at 14:14
• Thanks! Any idea of how to apply Chebyshev? – user817741 Dec 9 '20 at 14:14

By AM-GM, we have $$\sqrt{ab} + \sqrt{bc} + \sqrt{ca} \le a + b + c = 3$$. Thus, $$\mathrm{RHS} \ge \frac{1}{3}$$.
By Chebyshev's sum inequality, we have \begin{align} a^3 + b^3 + c^3 - (a+b+c) &= (a-1)(a^2 + a) + (b-1)(b^2+b) + (c-1)(c^2+c)\\ &\ge \frac{1}{3}(a-1 + b-1 + c-1)(a^2+a + b^2+b + c^2 + c)\\ & = 0. \end{align} Thus, we have $$a^3 + b^3 + c^3 \ge a + b + c = 3$$. Thus, we have \begin{align} a^6 + b^6 + 3c^3 + 4 &= (a^6 + 1) + (b^6 + 1) + 3c^3 + 2\\ &\ge 2a^3 + 2b^3 + 3c^3 + 2\\ &= 2(a^3 + b^3 + c^3) + c^3 + 2\\ &\ge 8 + c^3. \end{align} Thus, $$\mathrm{LHS} \le \sum_{\mathrm{cyc}} \frac{1}{8 + c^3}$$.
It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{1}{8 + c^3} \le \frac{1}{3}$$ or (tangent line method) $$\sum_{\mathrm{cyc}} \left(\frac{1}{8 + c^3} - \frac{1}{9} + \frac{1}{27}(c-1) \right) \le 0$$ or $$\sum_{\mathrm{cyc}} \frac{(c^2-2c-5)(c-1)^2}{27(8 + c^3)} \le 0$$ which is true since $$x^2-2x-5 < 0$$ for all $$0 \le x \le 3$$.
• very nice! What was your motivation when trying $\frac13\geqslant LHS$? – Dr. Mathva Dec 18 '20 at 16:25
• @Dr.Mathva Thanks. $c\mapsto \frac{1}{9} - \frac{1}{27}(c-1)$ is the tangent at $c=1$ of $c \mapsto \frac{1}{8+c^3}$. Then I tried $\frac{1}{8+c^3} \le \frac{1}{9} - \frac{1}{27}(c-1)$ for $0 \le c \le 3$ which is true. – River Li Dec 19 '20 at 1:09