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:
$$\left | 2x-y-4 \right |\geq 4\sqrt{2}+4$$
 A: $36 + 9y^2 = 4x^2$ Suggests we could use $y = 2 \tan A$ to get $x = \pm 3\sec A$.  Given the objective, it is sufficient to consider $x = 3 \sec A, ~~ A \in [0, \frac{\pi}{2})$
Then $(2x - y - 4 )^2 = 4(3 \sec A -  \tan A - 2)^2 $
It is easily checked that the minimum occurs when $3 \tan A = \sec A$ or when $\sin A = \frac{1}{3}$. So the minimum value is $4(3 \sec A -  \tan A - 2)^2 = 4(3\frac{3}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}-2)^2 = \left(4\sqrt{2}-4\right)^2$
A: Put $x=\pm3\cosh(t)$ and $y=2\sinh(t)$.
In the positive case, you obtain
$$f(t)=2x-y-4=6\cosh(t)-2\sinh(t)-4=2e^t+4e^{-t}-4=2e^{-t}(1+(e^t-1)^2)>0,$$
thus it's enought to find the minimum of $f$by solving $f'(t)=0$.
This gives $e^t=\sqrt 2$ from which $|f(t)|=f(t)\geq 4\sqrt 2-4$.
In the negative case, the lower bound is $|f(t)|=-f(t)\geq 4\sqrt 2+4$.
A: 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 $\left|2x-y-4\right| \geq 4\sqrt{2}-4$. $\square$
