If $a,b,c>0$ and $ab+bc+ca=3$, prove that $\sum_{cyc} \frac{a}{\sqrt{a^3+5}} \leq \sqrt{6}/2$

Question

If $a,b,c>0$ and $ab+bc+ca=3$, prove that $\displaystyle \sum_{cyc} \frac{a}{\sqrt{a^3+5}} \leq \sqrt{6}/2$.

My attempt was to use firstly AM-GM in the denominator, like $a^3+5 \geq 3a+3$ and the set $a=x^2-1$ but a lot of trouble occurred.

Then, I tried to use AM-HM, like $\sqrt{6 \cdot (a^3+5)} \geq ...$ but this didn't work either.

Any help?

I think it's completely obvious that if two problems have the same condition and are inequalities, are NOT the same.

Answer 1

Another AM-GM helps: $$\sum_{cyc}\frac{a}{\sqrt{a^3+5}}\leq\sum_{cyc}\frac{a}{\sqrt{\frac{3a^2-1}{2}+5}}=\sqrt{\frac{2}{3}}\sum_{cyc}\frac{a}{\sqrt{a^2+3}}=$$ $$=\sqrt{\frac{2}{3}}\sum_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}\leq\frac{1}{2}\sqrt{\frac{2}{3}}\sum_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{\sqrt6}{2}.$$