Prove $(x+y+z) \cdot \left( \frac1x +\frac1y +\frac1z\right) \geqslant 9 + \frac{4(x-y)^2}{xy+yz+zx}$ $x,y,z >0$, prove
$$(x+y+z) \cdot \left( \frac1x +\frac1y +\frac1z\right) \geqslant 9 + \frac{4(x-y)^2}{xy+yz+zx}$$
The term $\frac{4(x-y)^2}{xy+yz+zx}$ made this inequality tougher. It remains me of this inequality. I think one can prove it by expanding everything, but is there any elegant solution for this problem ?

There is a very nice solution only using SOS, I was found it!
We need to prove:$$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) - 9 \geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$(\frac{x}{y}+\frac{y}{x}-2) + (\frac{y}{z}+\frac{z}{x}-2)+(\frac{z}{x}+\frac{x}{z}-2) \geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$\frac{(x-y)^2}{xy}+\frac{(x-y)^2}{yz+zx}+(\frac{(y-z)^2}{yz}+\frac{(z-x)^2}{zx} -\frac{(x-y)^2}{yz+zx})\geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$(x-y)^2(\frac{1}{xy}+\frac{1}{yz+zx})+\frac{(xz+yz-2xy)^2}{xyz(x+y)} \geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$(x-y)^2(\frac{1}{xy}+\frac{1}{yz+zx}-\frac{4}{xy+yz+zx})+\frac{(xz+yz-2xy)^2}{xyz(x+y)} \geq 0$$
or $$\frac{(x-y)^2(xz+yz-xy)^2}{xyz(x+y)(xy+yz+zx)}+\frac{(xz+yz-2xy)^2}{xyz(x+y)} \geq 0$$
Which is obvious!
Equality holds when $x=y=z$
