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Prove that :- $$\frac{x^2}{y}+ \frac{y^2}{z}+\frac{z^2}{x} \geq x+y+z$$

Where $x,y,z$ are positive real numbers

My attempt :- L.H.S = $\frac{x^3 z +x y^3 + y z^3 }{xyz} $

We need to show that

$x^3 z + x y^3 + y z^3 \geq xyz (x+y+z) $

I tried the AM-GM

$3(x^3 z + x y^3 + y z^3) \geq (xyz)^{\frac{4}{3}}$

But i could not go on any more !

nonuser
  • 90,026
Hussien Mohamed
  • 1,802

1 Answers1

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We have by Cauchy-Schwarz-Bunyakovsky inequality: $$ (a^2+b^2+c^2)(x^2+y^2+z^2)\geq (ax+by+cz)^2$$ or writen like this: $$(a+b+c)(x+y+z)\geq (\sqrt{ax}+\sqrt{by}+\sqrt{cz})^2$$ we have a folloving: $$(y+z+x)(\frac{x^2}{y}+ \frac{y^2}{z}+\frac{z^2}{x}) \geq (x+y+z)^2$$ and thus a conclusion.

nonuser
  • 90,026