<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] To maximize $x^2+y^2+z^2$ subject to the constraint $\dfrac1x+\dfrac1y+\dfrac1z=1$, you could use the Lagrange multiplier method and look at $x^2+y^2+z^2-\lambda\left(\dfrac1x+\dfrac1y+\dfrac1z-1\right)$.  Take derivatives and see that $2x+\dfrac\lambda{x^2}=2y+\dfrac\lambda{y^2}=2z+\dfrac\lambda{z^2}=\dfrac1x+\dfrac1y+\dfrac1z-1=0,$ so $2x^3+\lambda=2y^3+\lambda=2z^3+\lambda,$ so $x=y=z$ and $x=3$ and $x^2+y^2+z^2=27$.
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