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.