Goutham wrote: Let $a, b$ and $c$ be positive real numbers. Prove that \[\frac{\sqrt{a^2+bc}}{b+c}+\frac{\sqrt{b^2+ca}}{c+a}+\frac{\sqrt{c^2+ab}}{a+b}\ge\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\] Let $a\geq b\geq c$. Hence, $\sum_{cyc}\left(\frac{\sqrt{a^2+bc}}{b+c}-\sqrt{\frac{a}{b+c}}\right)=$ $=\sum_{cyc}\frac{(a-b)(a-c)}{(b+c)\left(\sqrt{a^2+bc}+\sqrt{a(b+c)}\right)}\geq$ $\geq\frac{(a-b)(a-c)}{(b+c)\left(\sqrt{a^2+bc}+\sqrt{a(b+c)}\right)}+\frac{(b-a)(b-c)}{(a+c)\left(\sqrt{b^2+ac}+\sqrt{b(a+c)}\right)}=$ $=(a-b)\left(\frac{a-c}{(b+c)\left(\sqrt{a^2+bc}+\sqrt{a(b+c)}\right)}-\frac{b-c}{(a+c)\left(\sqrt{b^2+ac}+\sqrt{b(a+c)}\right)}\right)$, which is non-negative because $a\geq b$, $a-c\geq b-c$, $(a+c)\sqrt{b^2+ac}\geq(b+c)\sqrt{a^2+bc}$ and $(a+c)\sqrt{b(a+c)}\geq(b+c)\sqrt{a(b+c)}$.

Goutham wrote: Let $a, b$ and $c$ be positive real numbers. Prove that \[\frac{\sqrt{a^2+bc}}{b+c}+\frac{\sqrt{b^2+ca}}{c+a}+\frac{\sqrt{c^2+ab}}{a+b}\ge\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\] Because $\sum \frac{(a-b)(a-c)}{(b+c)^2} \ge 0$ $\Rightarrow \sum \frac{a^2+bc}{(b+c)^2} \ge \sum \frac{a}{b+c}$ and $\sum \frac{\sqrt{(a^2+bc)(b^2+ca)}}{(b+c)(c+a)}$ $=\sum \sqrt{\frac{(a^2+bc)(b+c)(b^2+ca)(a+c)}{(b+c)^3(c+a)^3}}$ $ \ge \sum \sqrt{\frac{b(a+c)^2a(b+c)^2}{(a+c)^3(b+c)^3}}$ $=\sum\sqrt{\frac{ab}{(a+c)(b+c)}} $ $\Rightarrow \left( \sum \frac{\sqrt{a^2+bc}}{b+c} \right)^2 \ge \left( \sum \sqrt{\frac{a}{b+c}}\right)^2 $