Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}}\le\sqrt[3]{9\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$$
It is from here.
Remark: At the end of the linked page arqady (a.k.a Michael Rozenberg) says "It seems that the following inequality is also true". So it might be wrong.
Obviously, $\sqrt[3]{9\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)}\ge3$. So it suffices to show that $\sum\sqrt{\dfrac{2a}{b+c}}\le3$, which is obviously wrong for $(a,b,c)=(10,1,1)$.
Or using Cauchy–Schwarz inequality we have:
$$\sum\sqrt{\dfrac{2a}{b+c}}\le\sqrt{2\left(a+b+c\right)\sum\frac{1}{a+b}}$$
So it suffices to show that $\sqrt{2\left(a+b+c\right)\sum\frac{1}{a+b}}\le\sqrt[3]{9\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$, which is wrong for $(a,b,c)=(1,2,3)$.