Prove that: $\left | 2x-y-4 \right |\geq 4\sqrt{2}+4$ Let $x,y\in \mathbb{R}$ know that $4x^2-9y^2=36$
Prove that:
$$\left | 2x-y-4 \right |\geq 4\sqrt{2}+4$$

Let $u=2x-3y$ and $v = 2x+3y$. Then $uv = 4x^2-9y^2 = 36$, so
$$2x-y-4 = {2u+v \over 3} - 4 = {2 \over 3}u + 12u^{-1} - 4$$
${\partial \over \partial u}$ of this is ${2 \over 3}-12u^{-2}$, which is zero at $u=\pm3\sqrt{2}$. Plugging these values into ${2 \over 3}u + 12u^{-1} - 4$ yields the local extrema $\pm4\sqrt{2}-4$. It follows that $\left|2x-y-4\right| \geq 4\sqrt{2}-4$. $\square$
