Inequality involving positive real numbers

[anemone]

Prove that $\dfrac{y^2z}{x}+y^2+z\ge\dfrac{9y^2z}{x+y^2+z}$ for all positive real numbers $x,\,y$ and $z$.

 

[lfdahl]

Spoiler: Suggested solution

I use the fact, that the arithmetic mean is always greater than or equal to the harmonic mean (for positive reals):

\[\frac{\frac{y^2z}{x}+y^2 +z}{3} \geq \frac{3}{\frac{x}{y^2z}+\frac{1}{y^2}+\frac{1}{z}}\]

It follows immediately, that

\[\frac{y^2z}{x}+y^2 +z \geq \frac{9y^2z}{x+z+y^2}\]

q.e.d.