<s>[INST] Maximizing $\sqrt6xy+4yz$ Let $x, y, z$ be real numbers such that $x^2+y^2+z^2=1$.
Let $A$ be maximum value of $\sqrt6xy+4yz$. Find $2A^2-4$.
An initial approach was trying out the inequality $\text{RMS-AM-GM}$. Then I tried a parametric substitution: $x=\cos\alpha, y=\sin\alpha cos\beta, z=\sin\beta \sin\alpha$.
How would you go about this?
 [/INST] As $y$ is repeated in  $\sqrt6xy+4yz,$
WLOG choose  $y=\cos\alpha, x=\sin\alpha\cos\beta, z=\sin\beta\sin\alpha$
$\sqrt6xy+4yz=\cos\alpha\sin\alpha(\sqrt6\cos\beta+4\sin\beta)$
$=\dfrac{\sin2\alpha}2\cdot\sqrt{22}\cos\left(\beta-\arccos\sqrt{\dfrac6{22}}\right)$
$\le\dfrac{\sqrt{22}}2$
The equality occurs if $\sin2\alpha=1$ and $\beta=\arccos\sqrt{\dfrac6{22}}$
$\iff\alpha=180^\circ n+45^\circ$ and $\cos\beta=\sqrt{\dfrac6{22}};\sin\beta=\dfrac4{\sqrt{22}}$
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