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

Question

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...

Thanks in advance!

Answer 1

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 $0 \le x \le 3$.