Minimum value?

Assume WLOG that both $x$ and $y$ are positive ( It doesn't matter since in all of the below equations $x^2, y^2, |x|, |y|$ all change the sign to positive. This implies we want to minimize $x-y$ .

Now, this implies $log_4(x+2y)+log_4(x-2y) = log_4((x+2y)(x-2y)) = log_4(x^2-4y^2) = 1$

Therefore, $x^2-4y^2=4^1 = 4$

Transposition gives us $x^2 = 4y^2+4$ , $x = \sqrt{4y^2+4}$ . But, by the Cauchy-Schwarz Inequality,

$(\frac{1}{2}^2+\frac{\sqrt{3}}{4}^2)((2y)^2+(2)^2 )= (4y^2+4) \geq (y+\sqrt{3})^2$ .

Therefore $x = \sqrt{4y^2+4} \geq \sqrt{(y+ \sqrt{3})^2} = y+ \sqrt{3}$

So, $x \geq y+\sqrt{3}$ , or $x-y \geq \sqrt{3}$ . So, our minimum is $\sqrt{3}$ . Our required answer is then $\lceil \sqrt{3} \rceil = \boxed{2}$ ,

Since $1<3<4$ and taking square roots of both sides gives us $1 < \sqrt{3} <2$ .

I'm going to take a calculus approach with LaGrange Multipliers. Let $f(x,y) = |x| - |y|$ and $g(x,y) = x^2 - 4y^2 = 4$ . Taking $grad(f) = \lambda \cdot grad(g)$ yields:

$\frac{x}{|x|} = \lambda (2x);$

$-\frac{y}{|y|} = \lambda (-8y);$

or $|x| = 4|y|.$ Substituting this value into $g(x,y)$ produces $16y^2 - 4y^2 = 4 \Rightarrow y^2 = \frac{1}{3} \Rightarrow |y| = \frac{1}{\sqrt{3}}.$ So we have the critical ordered-pair $(x,y) = (\frac{4}{\sqrt{3}}, \frac{1}{\sqrt{3}}),$ which $f(\frac{4}{\sqrt{3}},\frac{1}{\sqrt{3}}) = \sqrt{3}$ and the ceiling of which equals 2.