$x,y,z >0$, 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 ?


2 Answers 2


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: $$\frac{(x-z)^2}{xz}+\frac{(z-y)^2}{zy}+\frac{(y-x)^2}{xy}\geq \frac{\left(|x-z|+|z-y|+|y-x|\right)^2}{xy+xz+yz}. \tag{4}$$


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 $x=y=z$


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