<s>[INST] For $a,b,c>0$ and $a+b+c=6$. Prove that $\sum^{}_{cyc} \frac{ab}{\sqrt{a^2+b^2+2c^2}}\leq 3$ For $a,b,c>0$ and $a+b+c=6$. Prove that $$\sum^{}_{cyc} \frac{ab}{\sqrt{a^2+b^2+2c^2}}\leq 3$$
And this attempt $$\sum \frac{ab}{\sqrt{a^2+b^2+2c^2}}$$
$$=\sum \sqrt{\frac{a^2b^2}{\sqrt{a^2+b^2+2c^2}}}$$
$$\leq \sqrt{3}.\sqrt{\frac{(ab+ca+ca)^2}{4(a^2+b^2+c^2)}}$$
$$\leq \sqrt{3}.\sqrt{\frac{a^2+b^2+c^2}{4}}$$
 [/INST] From $(a+b+c)^2 \geqslant 3(ab+bc+ca),$ we get $ab+bc+ca \leqslant 12.$
Now, using the Cauchy-Schwarz and AM-GM inequality, we have
$$\left(\sum \frac{ab}{\sqrt{a^2+b^2+2c^2}}\right)^2  \leqslant (ab+bc+ca)\sum \frac{ab}{a^2+b^2+2c^2}$$
$$ \leqslant \frac{1}{4}(ab+bc+ca) \sum \frac{(a+b)^2}{a^2+b^2+2c^2}$$
$$ \leqslant \frac{1}{4}(ab+bc+ca) \sum \left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right) $$
$$ = \frac{3}{4}(ab+bc+ca) \leqslant  9.$$
Therefore
$$\sum \frac{ab}{\sqrt{a^2+b^2+2c^2}} \leqslant 3.$$
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