Prove that the minimum values of $x^2+y^2+z^2$ is $27$ with given condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$. Question: Prove that the minimum values of $x^2+y^2+z^2$ is $27$, where $x,y,z$ are positive real variables satisfying the condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$.  
From AM$\ge$ GM, we have  $\left( \dfrac{x^2+y^2+z^2}{3}\right)^3\ge (xyz)^2=(xy+yz+zx)^2$. Is it possible to show thae result from this relation?

This is a direct application of harmonic vs. quadratic mean (HM-QM):


*

*$\frac{3}{\frac 1x + \frac 1y + \frac 1z} \stackrel{HM-QM}{\leq}\sqrt{\frac{x^2+y^2+z^2}{3}}$
Now, plugin $\frac 1x + \frac 1y + \frac 1z = 1$ and you get
$$3 \leq \sqrt{\frac{x^2+y^2+z^2}{3}} \Leftrightarrow 27 \leq x^2+y^2+z^2$$
Equality holds for $x=y=z = 3$.
