<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 are trying to use GM<=AM, I suppose you have $x,y,z\geq 0$.
Check this link QM>=AM>=GM>=HM
First Step
Using HM<=GM:
$$ \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \leq \sqrt[3]{xyz} \implies xyz\geq \frac{3^3}{1}=27$$
Second Step
Using the AM<=QM:
$$ \frac{x+y+z}{3} \leq \sqrt{\frac{x^2+y^2+z^2}{3}}\implies \frac{x^2+y^2+z^2+2(xy+xz+yz)}{9}\leq \frac{x^2+y^2+z^2}{3}\implies \frac{x^2+y^2+z^2}{3}+\frac{2(xy+xz+yz)}{3}\leq x^2+y^2+z^2 \implies \frac{2}{3}(xy+xz+yz)\leq \frac{2}{3}(x^2+y^2+z^2)\implies xy+xz+yz\leq x^2+y^2+z^2$$
We know that $xy+xz+yz = xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=xyz$. Thus by First Step 
$$27\leq xyz \leq x^2+y^2+z^2$$
To finish this proof, note that $(x,y,z)=(3,3,3)$ is a feasible solution with an objective value equal to $27$.
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