Lagrange multiplier problem in book

Question

$f(x,y,z)=xyz$

$g(x,y,z)=x^2+2y^2+3z^2-6$

Then:

$\begin{cases}yz=2x\lambda\\xz=4y\lambda\\xy=6z\lambda\\x^2+2y^2+3z^2=6\end{cases}$

According to book's solution if any $x,y,z$ is equal to $0$, then $x=y=z=0$, which contradicts the last expression.

However, isn't possible to have $x=y=\lambda=0$ and $z\neq 0$ so last expression can be true?

Answer 1

The constraint must be binding, otherwise you could increase the objective function indefinitely. It follows that $\lambda \neq 0$, and then the argument works.