<s>[INST] 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?
 [/INST] If you wish to complete your AM-GM, note 
$$1=\frac1x+\frac1y+\frac1z\geqslant \frac3{\sqrt[3]{xyz}}\implies xyz\geqslant 27$$
Using that with what you got, viz.
$$\left(\frac{x^2+y^2+z^2}3 \right)^3 \geqslant (xyz)^2\geqslant 27^2=9^3 \implies x^2+y^2+z^2\geqslant 27$$
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