Obel1x wrote: If $a,b,c$ are real positive numbers prove that the inequality holds: \[ \frac{\sqrt{a^3+b^3}}{a^2+b^2}+\frac{\sqrt{b^3+c^3}}{b^2+c^2}+\frac{\sqrt{c^3+a^3}}{c^2+a^2} \ge \frac{6(ab+bc+ac)}{(a+b+c)\sqrt{(a+b)(b+c)(c+a)}} \] It is equal to: $\sum{}\frac{\sqrt{(a^3+b^3)(a+b)}}{a^2+b^2}\cdot\sqrt{(a+c)(b+c)}\ge \frac{6\sum{}ab}{\sum{}a}$ Now using Cauchy-Schwarz as follows we have: $\sqrt{(a^3+b^3)(a+b)}\ge a^2+b^2$ and $\sqrt{(a+c)(b+c)}\ge a+\sqrt{bc}$. Now, it remains to prove $(a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}) \cdot(a+b+c) \ge 6(ab+ac+bc)$ Or equally: $\sum{}(a^2+a\sqrt{ab}+a\sqrt{ac}+a\sqrt{bc}) \ge 4(ab+ac+bc)$ But using Schur inequality of degree 4 we have: $\sum{}(a^2+a\sqrt{ab}+a\sqrt{ac}+a\sqrt{bc}) \ge 2\sum{}(a\sqrt{ab}+b\sqrt{ab})$ And finally we need to prove just $\sum{}(a\sqrt{ab}+b\sqrt{ab}) \ge 2\sum{}ab$ what is equal to: $\sum{}\sqrt{ab}(\sqrt{a}-\sqrt{b})^2 \ge 0$ Done.