<s>[INST] 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 $y$.
 [/INST] We can use AM-GM inequality alone to settle the question. To this end, we find the constants $A,B,C,D$ such that: $\sqrt{6}xy \leq A^2x^2+B^2y^2$, and $4yz \leq C^2y^2+D^2z^2$ and $A^2 = B^2+C^2 = D^2, AB = \dfrac{\sqrt{6}}{2}, CD = 2 = AC$. Thus $A^4 = A^2\cdot A^2 = A^2(B^2+C^2) = (AB)^2 + (AC)^2 = \left(\dfrac{\sqrt{6}}{2}\right)^2 + 2^2 = \dfrac{11}{2}=\dfrac{22}{4}$. So $A^2 = \dfrac{\sqrt{22}}{2}$, and clearly this value of $A^2 = \dfrac{\sqrt{22}}{2}$ is the maximum we are looking for.
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