Problem 1-5 Trivial Inequality

Problem Posts

Problem 1-5 (Algebra) [Difficulty 2] [PDF]

QUESTION
(Sipnayan 2011 High School Elims, Average Question 2) Suppose $x$ and $y$ are real numbers such that $$2x^2+y^2-2xy+12y+72\le0$$ . What is the value of $x^2y$?

Overview/Laconic Solution Sketch
Group into squares and apply the Trivial Inequality.



Discussion
It's evident that Cauchy-Schwarz and AM-GM isn't very viable. We'll have to stick with good old Trivial Inequality ($x^2\ge 0, \forall x \in \mathbb{R}$) ($\forall$ is the universal quantifier)

The first sensible thing to do would be to try to group the left hand side into a sum of squares. If we tried guessing from the term $2xy$, it doesn't bring us far: $$\left(x^2-2xy+y^2\right) + x^2 + 12y + 72$$ The more annoying term turns out to be $12y$ and $72$. Now note that $72$ isn't a square, but rather half a square. Better yet, $2\cdot 72 = 144 = 12^2$. Hence, $12y+72 = \left( y+12 \right)^2 -\frac{y^2}{2}$. Now we'll find that $y^2/2$ is just what the neighboring expression needs.

Solution
$$\begin{eqnarray*} 2x^{2}+y^{2}-2xy+12y+72 & = & 2x^{2}-2xy+\frac{y^{2}}{2}+\frac{y^{2}}{2}+12y+72\\  & = & \frac{1}{2}\left(2x-y\right)^{2}+\frac{1}{2}\left(y+12\right)^{2}\\& \le&0 \end{eqnarray*}$$ which implies $y=-12$ and $x=-6$. We have $x^2y=\boxed{-432}$