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 $y$.

The easiest method seems to be to use Lagrange multipliers.
$$
L = \sqrt{6} xy + 4 yz - \frac{1}{2} \lambda (x^2 + y^2 + z^2 - 1) \, ,
$$
for some $\lambda$. Looking at derivatives
$$
\frac{\mathrm{d} L}{\mathrm{d} x} = \sqrt{6} y - \lambda x = 0
$$
$$
\frac{\mathrm{d} L}{\mathrm{d} y} = \sqrt{6} x + 4z - \lambda y = 0
$$
$$
\frac{\mathrm{d} L}{\mathrm{d} z} = 4y - \lambda z = 0
$$
Another equation is the original constraint $x^2 + y^2 + z^2 = 1$. Now one has to solve the set of equations. After a few steps one gets $\lambda^2 = 22$ and $x = \pm \sqrt{3/22}$. From that it follows that $y = \pm \sqrt{1/2}$ and $z = \pm \sqrt{4/11}$.
The signs of $x,y,z$ are not independent, but the maximum of $\sqrt{6}xy + 4 yz$ is at $(\sqrt{3/22},\sqrt{1/2},\sqrt{4/11})$ and it is $\sqrt{11/2}$.
