If $a,b,c,d>0$ prove $\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac{2c^2+da}{c+d+a}+\frac{2d^2+ab}{d+a+b} \geq a+b+c+d$ If $a,b,c,d>0$ prove
$$\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac{2c^2+da}{c+d+a}+\frac{2d^2+ab}{d+a+b} \geq a+b+c+d$$
I tried cauchy: $$\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac{2c^2+da}{c+d+a}+\frac{2d^2+ab}{‌​d+a+b}$$ with $$(a+b+c + b+c+d +c+d+a +d+a+b)$$ but I got $$\sqrt{2a^2+bc} + \sqrt{2b^2+cd}+\sqrt{2c^2+da}+\sqrt{2d^2+ab}$$. If without square root, I can get $$(a+b)^2+(b+c)^2+(c+d)^2+(d+a)^2$$ and I can apply p-norm.
Thanks

We need to prove that
$$\sum_{cyc}\frac{2a^2+bc}{a+b+c}\geq\sum_{cyc}a$$ or
$$\sum_{cyc}\left(\frac{2a^2+bc}{a+b+c}-a\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)(a-c)}{a+b+c}\geq0.$$
Now, easy to use BW.
Indeed, let $a=\min\{a,b,c,d\}$, $b=a+u$, $c=a+v$ and $d=a+w$ and the rest for you.
