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?

HINT
$$1=x^2+y^2+z^2=x^2+ay^2+(1-a)y^2+z^2 \ge 2\sqrt{a}xy+2\sqrt{(1-a)}yz \quad (\because \text{AM-GM})$$ 
$$a=\frac{3}{11}$$ 
Note that $2\sqrt{a}:2\sqrt{(1-a)}=\sqrt{6}:4$. 
Let $2\sqrt{a}= \sqrt{6}k$, and multiply $k$ on each side. 
Can you take it from here?
