Minimize the Expression 6

$\frac{x^2+y^2}{x-y} = x - y + \frac{2}{x-y} \geq 2\sqrt{2} = \sqrt{8} \text{ (AM - GM)}$

Equality holds when $x-y = \sqrt{2} , xy = 1$ These two equations are solvable. Here is some extra algebra practice, find the equality case!

Given: $\boxed{ xy =1}$ and $\boxed{ x > y}$

We need to first realize that any number $x$ , where $x>0$ , will satisfy the above where $y=\frac{1}{x}$
If $x$ is negative, $y$ would also need to be negative, but since $x > y$ , there is no $y$ which would support a negative value for $x$ .

How do we approach this? We can start off by plugging in different values for $x$ and $y$ , to get a sense of what our answer will look like. Remembering $y=\frac{1}{x}$ , we can write our equation like this:

Let $b = \sqrt{a}$
$f(x)= min\{\frac{x^2+x^{-2}}{x-x^{-1}}\} = b$

$f(2)= \frac{(2)^2+(\frac {1}{2})^2}{(2)-(\frac {1}{2})}=\frac{17}{6}=2.8\overline{3}$

Ok, now let's try to get a lower result:

$f(3) = \frac{(3)^2+(\frac {1}{3})^2}{(3)-(\frac {1}{3})}= \frac{41}{12} = 3.41\overline{6}$

Oops, wrong direction, let's return to our $f(2)$ and go lower. Since we know x cannot be 1, let's try going to the midpoint between 1 and 2 which is $1\frac{1}{2}$ .

$f(1\frac{1}{2}) = \frac{(\frac{3}{2})^2+(\frac{2}{3})^2}{(\frac{3}{2})-(\frac{2}{3})}= \frac{97}{30}=3.2\overline{3}$

Closer than $f(3)$ but our $f(2)$ was closer. Our 2 value seems to be about spot-on. And after fiddling with this problem for a while, since I'm not willing to throw in the towel when I solve problems like this all the time with my programmer skills, I decided since I've been brute forcing it already, let me brute force it with JavaScript's help, because I'm a programmer for crying out loud.

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var answer={x:null,a:null};
for(var i=1,a,x;i<=1000;i++)
{
    x=1+(i/1000);
    a=Math.pow((Math.pow(x,2)+Math.pow(1/x,2))/(x-(1/x)),2);
    if(answer.x===null||a<answer.a){answer.a=a;answer.x=x;}
}
console.log("x≈"+answer.x+"\na≈"+answer.a);

In chrome this yields:
x≈1.932
a≈8.000000141498505

And close enough to get the answer right, so I punched in $\boxed{8}$ and got it right.

If you want to know the value where $x$ is exactly 8, the answer is $\frac{\sqrt{3}+1}{\sqrt{2}}$ .
How did I get this answer? I solved Alan Yan's equation in the answers. I won't show my work for that though so you can test your knowledge. But you can validate it in JavaScript like so:

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var x=(Math.sqrt(3)+1)/Math.sqrt(2);
a=Math.pow((Math.pow(x,2)+Math.pow(1/x,2))/(x-(1/x)),2);
console.log(a);

Call it cheating if you will, I didn't give up and got it right. And to make up for it, not seeing an easy approach in the answers already, I posted this answer. Hope others find it useful.