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!
 A: 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 $\sum_{\text{cyc}} \sqrt{a^2+b^2} \ge \sum_{\text{cyc}} \dfrac{a+b}{\sqrt{2}}= \sum_{\text{cyc}} \sqrt{2}a$ .
A: Another way:
$$\sum_{cyc}\left(\frac{a^2}{b}+a-\sqrt{2(a^2+b^2)}\right)=\sum_{cyc}\left(\frac{a^2}{b}-2a+b-\left(\sqrt{2(a^2+b^2)}-a-b\right)\right)=$$
$$=\sum_{cyc}(a-b)^2\left(\frac{1}{b}-\frac{1}{\sqrt{2(a^2+b^2)}+a+b}\right)\geq0.$$
A: Also, we can use $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}$$ and
$$\sqrt{2(a^2+b^2)}\leq\frac{3a^2+2ab+3b^2}{2(a+b)}.$$
By this way we can get something stronger:

Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sqrt[3]{4(a^3+b^3)}+\sqrt[3]{4(b^3+c^3)}+\sqrt[3]{4(c^3+a^3)}.$$

The following inequality is also true.

Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sqrt[4]{8(a^4+b^4)}+\sqrt[4]{8(b^4+c^4)}+\sqrt[4]{8(c^4+a^4)}.$$

