For $a,b,c>0$ proving $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geqslant a + b + c + \frac{4(a - b)^2}{a + b + c}$

Question

The problem with which I have a problem it's this:

For $a,b,c>0$ prove that $$ \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geqslant a + b + c + \frac{4(a - b)^2}{a + b + c} $$

Titu's Lemma and AM-GM work no good because this similar looking inequality is sharper.
After trying these, I decided to go for the following.

Here is my work:
Multiply $abc(a + b + c)$ to both the sides, $$ (a + b + c) (a^3c + ab^3 + bc^3) \geqslant abc(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) + 4abc (a - b) ^2 $$ After some work we are left to prove that: $$ \sum_{cyc} {a^2b^3 + ab^4 - 2a^2 b^2 c} \geqslant abc(4 a^2 + 4 b^2 - 8ab) $$ How to prove this or there is some better approach?
This inequality was the first in a list of many inequalities, but I don't think that means it is easy.

Thanks for help!

Answer 1

By C-S $$\sum_{cyc}\frac{a^2}{b}-\sum_{cyc }a=\sum_{cyc}\left(\frac{a^2}{b}-2a+b\right)=\sum_{cyc}\frac{(a-b)^2}{b}\geq$$ $$\geq\frac{(a-b+c-b+a-c)^2}{a+b+c}=\frac{4(a-b)^2}{a+b+c}.$$

Answer 2

The SOS proof. We have $$\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} - (a + b + c) - \frac{4(a - b)^2}{a + b + c}$$ $$=\frac{a(b^2-2bc+ca)^2+b(c^2-ab)^2+c(a^2-2ab+bc)^2}{abc(a+b+c)} \geqslant 0.$$