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!

What this question is really asking is if the function has a global minimum at all, as it isn't even defined on the set $x=y$ as you mentioned.
The set $\{(x,y)\in R^2:x\ne y\}$ is open, and therefore ay local (and thus the global) minimum must have the partial derivatives equal $0$. The function is symmetric, so we assume WLOG $x>y$ and the function is now $H(x,y)=x^2+y^2+\frac{1}{x-y}$. Thus $H_x(x,y)=2x-\frac{1}{(x-y)^2}$, $H_y(x,y)=2y+\frac{1}{(x-y)^2}$. These must both be $0$, so the sum of them must be $0$, so $2x+2y=0$ so $x=-y$.
Plugging this back into $H_x$ and setting equal to $0$ gives $2x=\frac{1}{(2x)^2}$ so $8x^3=1$, so $x=\frac12$, and thus $y=-\frac12$. To see if this is actually a minimum, we find the Hessian at this point to be 
$$
\begin{bmatrix}
    2.5       & -.5  \\
    -.5       & 2.5  \\
\end{bmatrix}
$$
which has eigenvalues $2$ and $3$ so is positive definite, thus this is a local minimum. As they are the only local minima, the global minimum is achieved at $H(\frac12,-\frac12)=\frac32$, and by symmetry at $H(-\frac12,\frac12)=\frac32$ as well.
