HMMT-Nov 2018 guts, Problem 32

Over all real numbers $x$ and $y$, find the minimum possible value of \[ (x y)^{2}+(x+7)^{2}+(2 y+7)^{2} \]

Solution: Solution 1: Rewrite the given expression as $\left(x^{2}+4\right)\left(1+y^{2}\right)+14(x+2 y)+94$. By Cauchy-Schwartz, this is at least $(x+2 y)^{2}+14(x+2 y)+94=(x+2 y+7)^{2}+45$. The minimum is 45 , attained when $x y=2, x+2 y=-7$. Solution 2: Let $z=2 y, s=x+z, p=x z$. We seek to minimize \[ \begin{aligned} \left(\frac{x z}{2}\right)^{2}+(x+7)^{2}+(z+7)^{2} & =\frac{p^{2}}{4}+\left(x^{2}+z^{2}\right)+14(x+z)+98 \\ & =\frac{p^{2}}{4}+s^{2}-2 p+14 s+98 \\ & =\left(\frac{p}{2}-2\right)^{2}+(s+7)^{2}+45 \\ & \geq 45 \end{aligned} \] Equality holds when $s=-7, p=4$. Since $s^{2} \geq 4 p$, this system has a real solution for $x$ and $z$. $\fbox{45}$.

Answer: 45