2018 JBMO SL #A6

For $a,b,c$ positive real numbers such that $ab+bc+ca=3$, prove $$\frac{a}{\sqrt{a^3+5}}+\frac{b}{\sqrt{b^3+5}}+\frac{c}{\sqrt{c^3+5}} \leq \frac{\sqrt{6}}{2}$$

Basically, we have to prove that $$\sum_{cyc}\frac{a}{\sqrt{a^3+5}} \leq \frac{\sqrt{6}}{2}.$$

Firstly notice that $$2(a^3+5)=(a^3+a^3+1)+9\geq 3a^2+9=3(a+b)(a+c)$$. So, $$\sum_{cyc} \frac{a}{\sqrt{a^3+5}}\leq \sqrt{\frac{2}{3}}\sum_{cyc} \frac{a}{\sqrt{(a+b)(a+c)}}$$ $$=\sqrt{\frac{2}{3}\cdot \frac{\displaystyle{\left(\sum_{cyc} \sqrt{a^2(b+c)}\right)^2}}{(a+b)(b+c)(c+a)}}$$ (By Cauchy-Schwarz.)$$\leq \sqrt{\frac{2}{3}\cdot \frac{(\sum a)(\sum (ab+ac))}{(a+b)(b+c)(c+a))}}\leq \frac{\sqrt{6}}{2}$$