Show that the function $H(x, y) = x^2 + y^2 + |x − y|^{-1}$ achieves its global minimum somewhere on the set $\{(x, y) \in \mathbb{R}^2 : x \ne y\}$. Show that the function $H(x, y) = x^2 + y^2 + |x − y|^{-1}$ achieves its global minimum somewhere on the set $\{(x, y) \in \mathbb{R}^2 : x \ne y\}$.
I kind of understand the minimum cannot be $x=y$ since $1/(x-y)=1/0$, which is infinity. I am trying to prove it via contradiction, but I am confused about how to find the minimum of a function with two variables. 
Any advice or hints will be appreciated. Thank you!

You don't actually need to find the minimum to prove that there is in fact a global minimum. You can set the stage to use the Weierstrass theorem (any continuous function reaches a maximum and a minimum value over a compact—i.e., closed and bounded—set).
Here, a sketch of the proof, based on the idea that the minimum won't be very close to the line $x=y$ (where the values go to $+\infty$) nor very far from the origin (for the same reason).
So let's take any point in the domain, say $(1,0)$. Since $f(1,0)=2$, we know that if there is a minimum is not greater than $2$. 
Now, let's define a compact set such that on it's complement the function always takes values greater than $2$.
Define
$$A=\left\{(x,y)\colon |x-y|\ge\frac12\right\}$$
and
$$B=\left\{(x,y)\colon x^2+y^2\le2\right\}.$$
Is easy to see that the set $A\cap B$ is closed and bounded, that is, is compact, and since $f$ is continuous, reaches a minimum in this set. Also, this minimum value is no more than $2$.
And if $(x,y)$ is not in $A$ (that is $|x-y|<\frac12$) or if is not in $B$ (and so $x^2+y^2>2$), then is easy to prove that $f(x,y)>2$, and so the minimum reached in $A\cap B$ is a global minimum.
