Points on the surface $x y^3 z^2=16$ that are closest to origin

Question

I have to find the points on the surface

$$E = \{(x,y,z)\in\mathbb{R}^3 : xy^3z^2=16\}$$ that are closest to the origin. How should we approach to this problem?

Answer 1

Actually for finding $$ \max_{xy^3 z^2=16}\sqrt{x^2+y^2+z^2} $$ you do not need Lagrange multiplies, nor derivatives. By the AM-GM inequality $$ 256 = x^2 y^6 z^4 = 108\cdot x^2\cdot \frac{1}{3}y^2\cdot \frac{1}{3}y^2\cdot \frac{1}{3}y^2\cdot\frac{1}{2}z^2 \cdot\frac{1}{2}z^2 \leq 108\left(\frac{x^2+y^2+z^2}{6}\right)^6$$ hence $xy^3 z^2=16$ implies $$ \sqrt{x^2+y^2+z^2} \geq \sqrt[12]{\frac{256}{108}\cdot 6^6} = \color{red}{2\cdot 3^{1/4}} $$ with equality attained at the points such that $x^2=\frac{y^2}{3}=\frac{z^2}{2}$. Here it is a nice sketch of the situation:

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