<s>[INST] 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 ?
 [/INST] We may recall that Lagrange's identity gives a proof of the Cauchy-Schwarz inequality:
$$ (a^2+b^2+c^2)(d^2+e^2+f^2) = (ad+be+cf)^2 + (ae-bd)^2+(af-cd)^2+(bf-ce)^2 $$
and set $(a,b,c)=(\sqrt{x},\sqrt{y},\sqrt{z})$, $(d,e,f)=\left(\frac{1}{\sqrt{x}},\frac{1}{\sqrt y},\frac{1}{\sqrt z}\right)$ to get:
$$ (x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)= 9 + \sum_{cyc}\left(\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}}\right)^2 \tag{1} $$
so it is enough to prove that:
$$ \sum_{cyc}\frac{(x-y)^2}{xy}\geq 4\frac{(x-y)^2}{xy+xz+yz}\tag{2} $$
but that is a consequence of the CS inequality in the form:
$$ \frac{A^2}{a}+\frac{B^2}{b}+\frac{C^2}{c}\geq \frac{(A+B+C)^2}{a+b+c}\tag{3} $$
also known as Titu's lemma:
$$\frac{(x-z)^2}{xz}+\frac{(z-y)^2}{zy}+\frac{(y-x)^2}{xy}\geq \frac{\left(|x-z|+|z-y|+|y-x|\right)^2}{xy+xz+yz}. \tag{4}$$
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