$2x^2+ 3y^2+4z^2 =1$ find the maximum of $4x+3y+2z$ If $2x^2+ 3y^2+4z^2 =1$ find the maximum of $4x+3y+2z$.
This is a question from a regional math olympiad and thus there must exist solutions without application of calculus.
I have no idea how to begin.

For me,
$4x + 3y + 2z = \langle\begin{pmatrix} \sqrt{2} x \\ \sqrt{3}y \\ 2z\end{pmatrix}, \begin{pmatrix} 2\sqrt{2} \\\sqrt{3} \\\ 1\end{pmatrix}\rangle \leq 1\times 2\sqrt{3}=2\sqrt{3}$
