<s>[INST] Greatest value of $xz$ in expression 
Let $x,y,z,t\in\mathbb{R}$ and $x^2+y^2=9,z^2+t^2=4$ and $xt-yz=6$. Then greatest value of $xz$ is 

What i try
I am trying to solve without trigonometric substution
$$(xt-yz)^2+(xz+yt)^2=(x^2+y^2)(z^2+t^2)$$
$6^2+(xz+yt)^2=4\cdot 9\Longrightarrow (xz+yt)^2=0$
$$xz=-yt$$
How do i solve it Help me please
 [/INST] Notice that:
$$4(x^2+y^2) + 9(t^2 + z^2) = 12(xt - yz)$$
and this is equivalent with
$$(2x - 3t)^2+ (2y + 3z)^2 = 0$$
Thus $z=-\frac{2}{3}y$, and we want to maximize $-\frac{2}{3}xy$. This is simple, given that $x^2+y^2=9$ because:
$$0\leq (x+y)^2\Rightarrow -2xy \leq x^2+y^2=9$$
Therefore:
$$zx=-\frac{2}{3}xy\leq 3$$
with equality when $(x,z)=\left(\pm \frac{3}{\sqrt{2}},\pm \sqrt{2}\right)$.
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