I have a inequality, I don't know where to start Show that for $x,y,z > 0$ the inequality is true:
$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+x+y+z \geq \frac{(x+y)^2}{y+z}+\frac{(y+z)^2}{z+x}+\frac{(z+x)^2}{x+y}$
I have tried Holder, but i had no luck. Please give me a hint of how to start.
 A: The hint.
It's $$\sum_{cyc}\left(\frac{x^2}{y}-2x+y\right) \geq \sum_{cyc}\left(\frac{(y+z)^2}{z+x}-2(y+z)+(z+x)\right)$$ or
$$\sum_{cyc}\frac{(x-y)^2}{y}\geq\sum_{cyc}\frac{(x-y)^2}{z+x}$$  or
$$\sum_{cyc}(x-y)^2S_z\geq0,$$
where $S_z=\frac{1}{y}-\frac{1}{z+x}.$
Now, by C-S $$\sum_{cyc}S_x=\sum_{cyc}\left(\frac{1}{x}-\frac{1}{x+y}\right)=\sum_{cyc}\left(\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)-\frac{1}{x+y}\right)\geq$$
$$\geq\sum_{cyc}\left(\frac{1}{2}\cdot\frac{4}{x+y}-\frac{1}{x+y}\right)=\sum_{cyc}\frac{1}{x+y}>0$$ and
$$\sum_{cyc}S_xS_y=\sum_{cyc}\left(\frac{1}{z}-\frac{1}{x+y}\right)\left(\frac{1}{x}-\frac{1}{y+z}\right)=$$
$$=\sum_{cyc}\left(\frac{1}{xy}-\frac{1}{z(y+z)}-\frac{1}{x(x+y)}+\frac{1}{(x+y)(x+z)}\right)=$$
$$=\sum_{cyc}\left(\frac{1}{xy}-\frac{1}{y(x+y)}-\frac{1}{x(x+y)}+\frac{1}{(x+y)(x+z)}\right)=$$
$$=\sum_{cyc}\frac{1}{(x+y)(x+z)}>0.$$
Now, since $$\sum_{cyc}S_x>0,$$ we can assume that $S_y+S_z>0$.
Also, we have
$$\sum_{cyc}(x-y)^2S_z=(x-y)^2S_z+(x-y+y-z)^2S_y+(y-z)^2S_x=$$
$$(S_y+S_z)(x-y)^2+2(x-y)(y-z)S_y+(S_x+S_y)(y-z)^2$$ and it's enough to prove that
$$S_y^2-(S_z+S_y)(S_x+S_y)\leq0,$$
which is $\sum\limits_{cyc}S_xS_y\geq0$.
Done!
