Prove that $5x^2−2xy−8x+ 2y^2−2y+ 5 \ge 0$ for all $x, y\in\mathbb R$. When does equality occur? I tried grouping the $x$'s and the $y$'s but that didn't get me anywhere. I know that $5x^2, 2y^2$, and $5$ are always positive. I am not sure what to try next. 

By inspection, equality occurs for $x=y=1$. 
So the obvious choice is to group around these values. This also gives a somewhat more detailed explanation how you arrive at these magic squares:
$$
5x^2−2xy−8x+ 2y^2−2y+ 5 = \\
5((x-1) +1)^2−2((x-1)+1)((y-1) + 1)−8(x-1)+ 2((y-1)+1)^2−2(y-1) -5\\
= 5(x-1)^2  - 2(x-1)(y-1)  +2(y-1)^2 =\\
= 4(x-1)^2  + ((x-1)-(y-1))^2  +(y-1)^2 = \\
= 4(x-1)^2  + (x-y)^2  +(y-1)^2 \\
\geq 0
$$
which is clear.
