<s>[INST] $2x^2+ 3y^2+4z^2 =1$ 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.
 [/INST] WLOG, we can set $\sqrt2x=\cos A,\sqrt3y=\sin A\cos B,2z=\sin A\sin B$
$\implies4x+3y+2z=2\sqrt2\cos A+\sqrt3\sin A\cos B+\sin A\sin B$
$=2\sqrt2\cos A+\sin A(\sqrt3\cos B+\sin B)$
$=2\sqrt2\cos A+2\sin A\cos\left(B-\dfrac\pi6\right)$
If $\sin A\ge0,$
$4x+3y+2z\le2\sqrt2\cos A+2\sin A$ as $\cos\left(B-\dfrac\pi6\right)\le1$
The equality occurs if $\cos\left(B-\dfrac\pi6\right)=1$
$\iff B-\dfrac\pi6=2m\pi\iff B=2m\pi+\dfrac\pi6$ where $m$ is any integer
Again, $2\sqrt2\cos A+2\sin A=2(\sqrt2\cos A+\sin A)=2\sqrt3\sin\left(A+\arcsin\dfrac1{\sqrt3}\right)\le2\sqrt3$
The equality occurs if $\sin\left(A+\arcsin\dfrac1{\sqrt3}\right)=1$
$\iff A+\arcsin\dfrac1{\sqrt3}=2n\pi+\dfrac\pi2$
$\iff A=2n\pi+\dfrac\pi2-\arcsin\dfrac1{\sqrt3}=2n\pi+\arccos\dfrac1{\sqrt3}$
 where $n$ is any integer
Similarly, if $\sin A<0,$
$4x+3y+2z\le2\sqrt2\cos A+2\sin A$ as $\cos\left(B-\dfrac\pi6\right)\ge-1$
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