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?

Use lagrange multipliers. $\nabla f=\langle 2x,2y,2z\rangle$ and $\nabla g=\langle x^{-2},y^{-2},z^{-2}\rangle$ where $f(x,y,z)=x^2+y^2+z^2$ and $g(x,y,z)= x^{-1}+y^{-1}+z^{-1}$. Solving $\nabla f=\lambda\nabla g$ yields $\lambda=2x^3=2y^3=2z^3$ and therefore $x=y=z$. Plugging this into $g(x,y,z)=1$ we have $x=y=z=3$ which we plug into $f(x,y,z)$ to obtain $3^2+3^2+3^2=27$ which is obviously a minimum as $x^2+y^2+z^2$ is unbounded above on the surface $g(x,y,z)=1$.
