Agung wrote: Given $ a,b, c $ positive real numbers satisfying $ a+b+c=1 $. Prove that \[ \dfrac{1}{\sqrt{ab+bc+ca}}\ge \sqrt{\dfrac{2a}{3(b+c)}} +\sqrt{\dfrac{2b}{3(c+a)}} + \sqrt{\dfrac{2c}{3(a+b)}} \ge \sqrt{a} +\sqrt{b}+\sqrt{c} \] The left hand side can be easily proven by Jensen inequality otherwise we can use The AM-Gm inequality From Am_Gm we have $\sqrt{\frac{6a}{b+c}} \le \frac{3a}{2\sqrt{ab+bc+ca}}+\frac{\sqrt{ab+bc+ca}}{b+c}$ Thus $\sum \sqrt{\frac{6a}{b+c}} \le \sum \left(\frac{3a}{2\sqrt{ab+bc+ca}}+\frac{\sqrt{ab+bc+ca}}{b+c}\right)$ $=\frac{3}{2\sqrt{ab+bc+ca}}+\left(\sum \frac{1}{a+b}\right)\sqrt{ab+bc+ca}$ and it's easily seen that $\sum \frac{ab+bc+ca}{b+c}=\sum \left(a+\frac{bc}{b+c}\right)$ $\le \sum \left(a+\frac{b+c}{4} \right)=\frac{3}{2}$ Thus $\frac{3}{2\sqrt{ab+bc+ca}}+\left(\sum \frac{1}{a+b}\right)\sqrt{ab+bc+ca}$ $\le\frac{3}{2\sqrt{ab+bc+ca}}+\frac{3}{2\sqrt{ab+bc+ca}}=\frac{3}{\sqrt{ab+bc+ca}} $ and hence the result For the right hand side Am-Gm kills it $\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2a}{b+c}}+\frac{3\sqrt{3a}(b+c)}{2}\ge 3\sqrt[3]{\left(\sqrt{\frac{2a}{b+c}}\right)^2\frac{3\sqrt{3a}(b+c)}{2}}=3\sqrt{3a}$ Hence it suffices to Prove that $\sum \left(\frac{3\sqrt{3a}-\frac{3\sqrt{3a}(b+c)}{2}}{2}\right) \ge \sum \sqrt{3a}$ which reduces to $\sum \sqrt{3a} \ge \sum \frac{3\sqrt{3a}(1-a)}{2}$ Or $\sum 3a \sqrt{a} \ge \sum \sqrt{a}$ which is obviously true The proof is completed equality occurs when $a=b=c=\frac{1}{3}$

left just jensen right hölder,am-gm