How prove this inequality $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y}\ge x^2+y^2+z^2$ let $x\ge y\ge z\ge 0$,show that
$$\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge x^2+y^2+z^2$$
my try:
$$\Longleftrightarrow x^3y^2+y^3z^2+z^3x^2\ge xyz(x^2+y^2+z^2)$$
 A: let $x=z+u,y=z+v,\to u\ge v \ge0$
$x^3y^2+y^3z^2+z^3x^2- xyz(x^2+y^2+z^2)=(u^2-uv+v^2)z^3+3u^2vz^2+uv(u^2-v^2)z+3u^2v^2z+u^3v^2 \ge0$
A: let $$E(a,b,c)=a^3b^2+b^3c^2+c^3a^2-abc(a^2+b^2+c^2)$$
then
$$2E(a,b,c)=\sum a^3(b-c)^2-\sum a^2(b^3-c^3)=\sum a^2(b-c)^2(a+c-b)\ge 0$$
because
$$\sum a^2(b^3-c^3)=\sum a^2(b-c)^3$$
A: The equation can be rewritten as:
$$x^2 \left(\frac y z - 1\right) + y^2 \left(\frac z x - 1\right) + z^2 \left(\frac x y - 1\right) > 0 $$
Then since I noticed that $x > 0, y > 0, z > 0$ is not sufficient to solve the problem, I did a change of variables to convert the problem into one of that form:
$$v = \frac x y - 1, v > 0$$
$$u = \frac y z - 1, u > 0$$
$$\frac {x^2} {y^2} \left(\frac y z - 1\right) + \left(\frac z x - 1\right) + \frac {z^2} {y^2} \left(\frac x y - 1\right) > 0 $$
$$(v+1)^2u + (v+1)^{-1}(u+1)^{-1} - 1 + (u + 1)^{-2}\left((v+1)(u+1)-1\right) > 0$$
Then I put it into a computer to simplify the fraction because no way I want to do that by hand...
$$\frac{\left( {u}^{3}+2\,{u}^{2}+u\right) \,{v}^{3}+\left( 3\,{u}^{3}+6\,{u}^{2}+4\,u+1\right) \,{v}^{2}+\left( 3\,{u}^{3}+5\,{u}^{2}+3\,u\right) \,v+{u}^{3}+{u}^{2}+u}{\left( {u}^{2}+2\,u+1\right) \,v+{u}^{2}+2\,u+1} > 0$$
That's my first time solving this kind of problem.
A: Another way.
$$\sum_{cyc}\left(\frac{x^2y}{z}-x^2\right)=(y-z)^2\left(\frac{x}{z}+\frac{x}{y}-1\right)+(x-y)(x-z)\left(\frac{y}{z}+\frac{y}{x}-1\right)\geq0.$$
