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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!

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I find the question hard to read. The way it looks, we have $2^2-3xy +2y^2$. This is negative at for example $(x,y)=(3,1)$, and many other places. But my guess is that the "$x$" is really supposed to be $\times$, and you are being asked about $2^2-3y+2y^2$. To tackle this, you could complete the square. But I do not want to give details until the question is clarified. –  André Nicolas Sep 19 '11 at 6:54
    
Sorry, should have been 2x^2. My bad! And sorry for not knowing how to make everything look pretty either... I looked in the FAQ and couldn't find the syntax for it. Didn't know where else to look. –  Matt Nashra Sep 19 '11 at 6:58
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Matt: You can learn about the LaTeX markup language by seeing Wikipedia or playing around in codecogs' sandbox. You can right click > "show source" any equation on Math.SE to view how the author typeset the equation. Lastly, here we put markup in between dollar signs, e.g. $\LaTeX\to$$\to\LaTeX$ (or two dollar signs for a displaystyle equation). –  anon Sep 19 '11 at 7:05
    
Also, do you know about a technique called "completing the square"? If you interpret the $y$ as a constant and the expression as a polynomial in $x$ then you might be able to c.t.s. and get a viable expression in the end... –  anon Sep 19 '11 at 7:08
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Just compute the discriminant of $ax^2 + bx + c=0$ with $a=2,b=-3y,c=2y^2$ –  Robert William Hanks Sep 19 '11 at 10:17
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4 Answers 4

up vote 10 down vote accepted

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.

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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.

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Nice solution, preserving the original symmetry. –  André Nicolas Sep 19 '11 at 12:37
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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?

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You don't really need to factor anything, just compute the discriminant of $ax^2 + bx + c=0$ with $a=2,b=-3y,c=2y^2$ –  Robert William Hanks Sep 19 '11 at 10:16
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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 $2x^2-3xy+2y^2 \ge 0$.

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