What is the maximum value of $\sqrt6 xy + 4yz$ given $x^2 + y^2 + z^2 = 1?$ Problem:
Let $x$,$y$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 1.$ Find the maximum possible value of $\sqrt6 xy + 4yz$

I don't know how to proceed with the question. Applying AM-GM inequality doesn't work because the second equation gives RHS dependent on $y$.

Let $a,b\in\mathbb{R}$.  We shall maximize $axy+byz$ subject to $x^2+y^2+z^2=1$ ($x,y,z\in\mathbb{R}$).
By Cauchy-Schwarz, $axy+byz=(ax+bz)y\leq \left(\sqrt{a^2+b^2}\sqrt{x^2+z^2}\right)|y|$.  By AM-GM, $\sqrt{x^2+z^2}\,|y|\leq \frac{\left(x^2+z^2\right)+y^2}{2}=\frac{1}{2}$.  That is, $axy+byz\leq \frac{\sqrt{a^2+b^2}}{2}$.  The equality holds iff $\left(x,y,z\right)=\pm\left(\frac{a}{\sqrt{2\left(a^2+b^2\right)}},\frac{1}{\sqrt{2}},\frac{b}{\sqrt{2\left(a^2+b^2\right)}}\right)$.  
(Similarly, we also have the lower bound $axy+byz\geq -\frac{\sqrt{a^2+b^2}}{2}$.  The equality holds iff $\left(x,y,z\right)=\pm\left(\frac{a}{\sqrt{2\left(a^2+b^2\right)}},-\frac{1}{\sqrt{2}},\frac{b}{\sqrt{2\left(a^2+b^2\right)}}\right)$.)
