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?
 A: 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.

A: Since another user has given the essential hint of this problem. Let's us look at a generalization of the problem and spell out some details.
For any $a,b,c > 0$, consider the function
$$G(x,y) = x^2y^2 + (ax+c)^2 + (by+c)^2$$
Expanding RHS out, we get
$$\begin{align}G(x,y) 
&= x^2y^2 + a^2x^2 + b^2y^2 + 2c(ax+by) + 2c^2\\
&= (x^2+b^2)(y^2 + a^2) + 2c(ax+by) + 2c^2 - a^2b^2
\end{align}$$
Notice
$$(x^2+b^2)(y^2+a^2) = |(x+bi)(y+ai)|^2 = |(xy-ab) + (ax+by)i|$$
We find
$$\begin{align}G(x,y)
&= (xy-ab)^2 + (ax+by)^2 + 2c(ax+by) + 2c^2 - a^2b^2\\
&= (xy - ab)^2 + (ax+by+c)^2 + c^2 - a^2b^2
\end{align}
$$
This implies $G(x,y) \ge c^2 - a^2b^2$ for all $(x,y) \in \mathbb{R}^2$.
The locus of $xy-ab$ is a hyperbola. If one look at its lower-left branch,
it has negative $x$ and negative $y$ axis as asymptotes. It is clear
the two "ends" of this branch lies below the line $ax+by+c = 0$.
The point $(-b,-a)$ belong to this branch. Since $ax + by + c$ evaluates
to $c-2ab$ at this point. If $c > 2ab$, this point lies above the line $ax+by+c = 0$.
In this case, the hyperbola $xy-ab$ intersect the line $ax+by+c = 0$ at two points
$(t_1,\frac{ab}{t_1})$, $(t_2,\frac{ab}{t_2})$ for some $t_1 < -b < t_2 < 0$.
In short, when $c > 2ab$, $G(x,y)$ achieves the value $c^2-a^2b^2$ at some point.
As a result, we have proved that

Given $a,b,c > 0$. If $c > 2ab$, then
  $$\min_{(x,y)\in\mathbb{R}^2} \left( x^2y^2 + (ax+c)^2 + (by+c)^2 \right) = c^2 - a^2b^2$$

For the function at hand, $(a,b,c) = (1,2,7)$ and $c > 2ab$. The minimum we seek is
$7^2 - 1^2 2^2 = 45$.
