<s>[INST] What's the approach for this inequality question? The question:
$a,b,c > 0; ab+bc+ca =3$, Prove that
$\sum_{cyc}\frac{a}{\sqrt{a^{3}+5}}\leq\frac{\sqrt{6}}{2}$
The sum is cyclic over $a,b,c$
I've looked at the problem for a long time but still can't think of an approach for this, so how can I solve this?
 [/INST] We have
\begin{align*}
 \sum_{\mathrm{cyc}} \frac{a}{\sqrt{a^3 + 5}}
 &= \sum_{\mathrm{cyc}} \frac{a}{\sqrt{\frac{a^3}{2} + \frac{a^3}{2} + \frac12 + \frac92}}\\
 &\le \sum_{\mathrm{cyc}} \frac{a}{\sqrt{3\sqrt[3]{\frac{a^3}{2} \cdot \frac{a^3}{2} \cdot \frac12} + \frac92}}\\
 &= \sum_{\mathrm{cyc}} \frac{2a}{\sqrt{6a^2 + 18}}\\
 &= \sum_{\mathrm{cyc}} \frac{2a}{\sqrt{6a^2 + 6(ab + bc + ca)}}\\
 &= \frac{1}{\sqrt{6}}\sum_{\mathrm{cyc}} 2\sqrt{\frac{a}{a + b}}  
 \sqrt{\frac{a}{a + c}}\\
 &\le \frac{1}{\sqrt{6}}\sum_{\mathrm{cyc}}
 \left(\frac{a}{a + b} + \frac{a}{a + c}\right)\\
 &= \frac{\sqrt{6}}{2}.
\end{align*}
We are done.
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