Find infimum and supremum of the function $f(x,y,z)=xyz$ on the unit sphere

Question

Find infimum and supremum of the function $ f (x, y, z) = xyz $ on the unit sphere $\{ (x,y,z) \in \mathbb R^3: x^2+y^2+z^2=1 \}$

From the Lagrange theorem: if the infimum and supremum exist then: \begin{align} yz&=\alpha x \tag{1}\\ xz&=\alpha y\tag{2} \\ xy&=\alpha z \tag{3}\\ x^2+y^2+z^2&=1 \tag{4}\end{align}

However, this system of equations seems to have many solutions, and I don't know how to find them cleverly.

Answer 1

By A-G inequality: $\sqrt[3]{|xyz|}\leq \sqrt{\frac{x^2+y^2+z^2}{3}}=\frac{1}{\sqrt{3}}$, therefore $-\frac{1}{3\sqrt{3}}\leq xyz\leq\frac{1}{3\sqrt{3}}$. Minimum is attained when $x=y=z=-\frac{1}{\sqrt{3}}$ and maximum for $x=y=z=\frac{1}{\sqrt{3}}$.

Answer 2

HINT

Based on the method proposed, suppose that $xyz\neq 0$.

Then one gets the following system of equations: \begin{align*} \begin{cases} xyz = \alpha x^{2}\\\\ xyz = \alpha y^{2}\\\\ xyz = \alpha z^{2} \end{cases} \Rightarrow x^{2} + y^{2} + z^{2} = \frac{3xyz}{\alpha} = 1 \Rightarrow xyz = \frac{\alpha}{3} \end{align*}

Consequently, one concludes that \begin{align*} (x,y,z) = \left(\pm\frac{1}{\sqrt{3}},\pm\frac{1}{\sqrt{3}},\pm\frac{1}{\sqrt{3}}\right) \end{align*}

Can you take it from here?