I have a inequality, I don't know where to start

Question

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.

Answer 1

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!