<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] You can still finish it off if you notice that $( xy+yz+zx )^2 = (xyz)^2 = xy\cdot yz \cdot zx \le \dfrac{(xy+yz+zx)^3}{27} \implies xy+yz+zx \ge 27 \implies \dfrac{x^2+y^2+z^2}{3} \ge \sqrt[3]{27^2} = 9 \implies x^2+y^2+z^2 \ge 9\cdot 3 = 27 $.
</s>