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?
On
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 $7^2 - 1^2 2^2 = 45$.
Hint:
\begin{align} f(x,y) &= (xy)^2+(x+7)^2+(2y+7)^2 \\ &= (xy-2)^2+(x+2y+7)^2+45 \\ \end{align}