Prove that for any $a,b,c>0$ we have: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\frac{(a+b)^2}{b+c}+\frac{(b+c)^2}{c+a}+\frac{(c+a)^2}{a+b}$$
Can someone help me? I tried Radon inequality for $p=1$, and I obtained $$2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge\frac{(a+b)^2}{b+c}+\frac{(b+c)^2}{c+a}+\frac{(c+a)^2}{a+b}$$ but that's the far I could go.