How to prove:$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sum_{cyc}{\sqrt{2(a^2+b^2)}}$ Problem: Let $a,b,c>0. $ Prove that: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sqrt{2(a^2+b^2)}+\sqrt{2(b^2+c^2)}+\sqrt{2(c^2+a^2)}$$
I have seen problem before, and I tried to prove: $$2(a+b+c)\ge\sqrt{2(a^2+b^2)}+\sqrt{2(b^2+c^2)}+\sqrt{2(c^2+a^2)}(!)$$
since by C-S inequality: $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c$. But (!) is not true.
Anyone can help me give a hint to solve this nice problem?. Thanks!

Applying Cauchy-Schwarz inequality:  $LHS = \sum_{\text{cyc}} \dfrac{a^2+b^2}{b}=\sum_{\text{cyc}}\dfrac{\left(\sqrt{a^2+b^2}\right)^2}{b}\ge\dfrac{\left(\sum_{\text{cyc}}\sqrt{a^2+b^2}\right)^2}{\sum_{\text{cyc}} a}\ge RHS\iff \sum_{\text{cyc}}\sqrt{a^2+b^2}\ge \sum_{\text{cyc}} \sqrt{2}a$, but this is true since $\sum_{\text{cyc}} \sqrt{a^2+b^2} \ge \sum_{\text{cyc}} \dfrac{a+b}{\sqrt{2}}= \sum_{\text{cyc}} \sqrt{2}a$ .
