How Can You Minimize |x|-|y| Given Logarithmic Constraints?

[anemone]

Here is this week's POTW:

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Minimize $|x|-|y|$, given $$\log_4{(x+2y)}+\log_4{(x-2y)}=1$$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to greg1313 for his correct solution, and you can find the suggested solution below::)

Spoiler

From the nature of the logarithm functions, we have:
$x+2y>0$ and $x-2y>0 \implies x>2|y|\ge 0$

We also know that $(x+2y)(x-2y)=4\implies x^2-4y^2=4$

By the symmetry, there is no loss of generality in considering only the case where $y\ge 0$.

In view of $x>0$, we need only to figure out the minimum value of $x-y$.

Letting $k=x-y$ and substituting it into $x^2-4y^2=4$, we get:

$3y^2-2ky+4-k^2=0$ for which it has real solutions, so we have:

$(-2k)^2-4(3)(4-k^2)\ge 0$

$k\ge \sqrt{3}$

If $k=\sqrt{3}$, we get $x=\dfrac{4\sqrt{3}}{3}$ and $y=\dfrac{\sqrt{3}}{3}$.

$\therefore$ the minimum of $|x|-|y|=\sqrt{3}$.

Alternative solution by greg1313:

Spoiler

$$\log_4(x-2y)+\log_4(x+2y)=1\implies x^2-4y^2=4\implies\frac{x^2}{4}-y^2=1$$

Now, we wish to minimize $|x|-|y|$ subject to $\frac{x^2}{4}-y^2=1$.
Note that this equation is an east-west hyperbola so, WLOG,
we need only address solutions in the first quadrant, where
$|x|=x$ and $|y|=y$.

Employing Lagrange multipliers,

$$x-y+\lambda\left(\frac{x^2}{4}-y^2-1\right)$$

so we wish to solve the system of equations

$$1+\frac{\lambda x}{2}=0$$

$$-1-2\lambda y=0$$

$$\frac{x^2}{4}-y^2-1=0$$

for $x$, $y$ and $\lambda$. Solving $x$ and $y$ for $\lambda$ in the first two equations
and substituting the results into the third equation gives us $\lambda=\pm\frac{\sqrt3}{2}$.
Then $x=\pm\frac{4\sqrt3}{3}$ and $y=\pm\frac{\sqrt3}{3}$, so

$$\min(|x|-|y|)=\sqrt3$$