Prove an inequality for positive real numbers 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 !

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.
