<s>[INST] 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...
Thanks in advance!
 [/INST] 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 $0 \le x \le 3$.
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