Minimize the expression This was a problem given in a no calculator math contest (HMMT):
What is the minimum value of $(xy)^2+(x+7)^2+(2y+7)^2$ where $x$ and $y$ are reals. My friends and I discussed about using calculus, but we reasoned that there must be a faster trick (calc would take too long). Any ideas?

Hint:
\begin{align}
  f(x,y) &= (xy)^2+(x+7)^2+(2y+7)^2 \\
  &= (xy-2)^2+(x+2y+7)^2+45 \\
\end{align}

Details for the thoughts
By $(a+b+c)^2=a^2+b^2+c^2+2(bc+ca+ab)$
Hence, 
  \begin{align}
  (x+7)^2+(2y+7)^2 &= x^2+14x+49+4y^2+28y+49 \\
  &= (x^2+4y^2+49+14x+28y \color{red}{+4xy})+49 \color{red}{-4xy} \\
  &= (x+2y+7)^2+49-4xy \\
 \end{align}
Also, 
  $$(xy)^2-4xy+4=(xy-2)^2$$
Therefore,
\begin{align}
f(x,y) &= (xy-2)^2+(x+2y+y)^2+49-4 \\
 &= (xy-2)^2+(x+2y+7)^2+45
\end{align}
Note that $(xy-2)^2 \ge 0 \, $ and $\, (x+2y+7)^2 \ge 0$, so $f(x,y)$ attains minimum when
$$
\left \{
  \begin{align}
    xy-2 &= 0 \\
    x+2y+7 &= 0
 \end{align}
\right.$$
which gives real solutions.

