$2x^2+ 3y^2+4z^2 =1$ 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.

Apply Cauchy-Schwarz Inequality: $\left(4x+3y+2z\right)^2 = \left(2\sqrt{2}\cdot \sqrt{2}x+\sqrt{3}\cdot \sqrt{3}y+1\cdot 2z\right)^2 \leq \left((2\sqrt{2})^2+(\sqrt{3})^2+1^2\right)\cdot\left(2x^2+3y^2+4z^2\right) = 12 \Rightarrow 4x+3y+2z \leq 2\sqrt{3} = \text{Maximum Value}$.
