Probar

Question

Demuestre:$$\frac{2x^2+xy}{(y+\sqrt{zx}+z)^2}+\frac{2y^2+yz}{(z+\sqrt{xy}+x)^2}+\frac{2z^2+zx}{(x+\sqrt{yz}+y)^2}\ge1$$ ($ x, y, z> 0 $)

Answer 1

Usando Cauchy-Schwarz, para cada expresión:

$\left ( y+\sqrt{xz}+z \right )^{2}\le \left ( y+x+x \right )\left ( y+z+\frac{z^{2}}{x} \right )=\dfrac{\left ( 2x^{2}+xy \right )\left ( xy+xz+z^{2} \right )}{x^{2}}$

Tenemos,

$\dfrac{2x^2+xy}{(y+\sqrt{zx}+z)^2}+\dfrac{2y^2+yz}{(z+\sqrt{xy}+x)^2}+\dfrac{2z^2+zx}{(x+\sqrt{yz}+y)^2}\ge \sum_{cyclic} \dfrac{x^{2}}{xy+xz+z^{2}}$

y otra vez usando Cauchy-Schwarz:

$\sum_{cyclic} \dfrac{x^2}{xy+xz+z^2}\ge \dfrac{(x+y+z)^2}{2(xy+yz+zx)+x^2+y^2+z^2}=\dfrac{(x+y+z)^2}{(x+y+z)^2}=1$. QED