Prove that $2x^2 - 3xy + 2y^2 \geq 0$ 
Prove that $2x^2 - 3xy + 2y^2 \geq 0$.

This is a question on my homework assignment, but I don't even know where to begin as it is not factorable and that is my first instinct when I see this type of problem. Can I get a tip on where to begin at least? Thanks!
 A: If you could show this is the sum of squares then you would be done.  For example, you know $x^2-2xy+y^2= (x-y)^2 \ge 0$.
The problem term in the question's expression is $-3xy$, which looks a little like the $-2xy$ I just mentioned.  So separate out $\frac{3}{2}$ times the expression above, to give $$\frac{3}{2}(x^2-2xy+y^2) + \frac{1}{2} x^2 +\frac{1}{2} y^2$$
and since this is the sum of positive fractions of squares, it is non-negative.     
A: We are asked to show that $2x^2-3xy+2y^2 \ge 0$, presumably for all real $x$ and $y$.
The standard approach is to complete the square.  We will do it in an ugly mechanical way.  Note that
$$2x^2-3xy+2y^2=2\left(x^2-\frac{3}{2}xy+y^2\right).$$
So it is enough to show that $x^2-\frac{3}{2}xy+y^2 \ge 0.$ Complete the square. We get
$$x^2-\frac{3}{2}xy+y^2=\left(x-\frac{3}{4}y\right)^2 -\frac{9}{16}y^2+y^2=\left(x-\frac{3}{4}y\right)^2 +\frac{7}{16}y^2.$$
Now we are finished. The expression on the right is obviously non-negative, since both $(x-(3/4)y)^2$ and $(7/16)y^2$ are non-negative. Indeed, "almost always" the expression is $>0$.  The only way it can be $0$ is if both $y$ and $x-(3/4)y$ are $0$, that is, if $x$ and $y$ are both $0$.
Comment: This looks like a "two-variable" problem, but it really isn't. Note that our inequality is clearly true if $y=0$.  So from now on we can assume that $y \ne 0$.  For $y \ne 0$, our inequality is equivalent to 
$$\frac{2x^2-3xy+2y^2}{y^2} \ge 0,$$
which in turn is equivalent to showing that
$$2z^2-3z+2 \ge 0,$$
where $z=x/y$.  Now we are down to a one-variable problem.  One standard approach is (again) by completing the square, but there are other ways to tackle the problem. For instance, note that $2z^2-3z+2$ is certainly $>0$ sometimes.  In order to be $<0$ sometimes, it would have to be $0$ for some $z$.  But it is easy to verify using the Quadratic Formula that the equation $2z^2-3z+2=0$ has no real solutions. 
A: Factor $2y^2$ in order to get $E(x,y)=2x^2-3xy+2y^2=2y^2((\frac{x}{y})^2-\frac{3}{2}\frac{x}{y}+1)$. The case $y=0$ is solved because $E(x,0)=2x^2\geq 0$. The problem is reduced to determining the sign of $F(t)=t^2-\frac{3}{2}t+1$ avec $t=\frac{x}{y}$. What is the discriminant of F?
A: Let's consider two cases, (i) $xy<0$, (ii) $xy\ge0$.
(i) In this case $-3xy>0$, and since $x^2, y^2 > 0$, we have $2x^2-3xy+y^2 > 0$.
(ii) In this case $(2x^2-3xy+2y^2) \ge (2x^2-3xy+2y^2)-xy =  2(x^2-2xy+y^2) = 2(x-y)^2 \ge 0$.
As cases (i) and (ii) are exhaustive (for real x,y), we have $2x^2-3xy+2y^2 \ge 0$.
