8
$\begingroup$

I came across the following problem that says:

If $x \neq 0,y \neq 0,$ then $x^2+xy+y^2$ is
1.Always positive
2.Always negative
3.zero
4.Sometimes positive and sometimes negative.

I have to determine which of the aforementioned options is right.

Now since $x \neq 0,y \neq 0$, so $ x^2+xy+y^2=(x-y)^2+3xy > 0$,if $x,y$ are of same sign. But if $x,y$ are of different sign,I am not sure about the conclusion.

Can someone point me in the right direction? Thanks in advance for your time.

$\endgroup$
  • 1
    $\begingroup$ 2. Because $x^2+xy+y^2=x^2(1+(y/x)+(y/x)^2)=x^2[(y/x + 1/2)^2 +3/4]>0$ $\endgroup$ – Hanul Jeon Apr 3 '13 at 9:38
  • $\begingroup$ You can check the sign of discriminant and conclude . $\endgroup$ – jdoicj Apr 3 '13 at 12:54
37
$\begingroup$

$x^2+xy+y^2 = \frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2 >0$, when $x \ne 0$ and $y \ne 0$.

$\endgroup$
  • 2
    $\begingroup$ Lovely solution! (+5 if I could) $\endgroup$ – Ittay Weiss Apr 3 '13 at 9:38
  • 3
    $\begingroup$ Why did you say "when $x\neq0$ and $y\neq0$"? You need neither for the equality, and only one of them for the inequality. $\endgroup$ – Marc van Leeuwen Apr 3 '13 at 9:54
  • 1
    $\begingroup$ @MarcvanLeeuwen, yes, answer can be expanded with words "equality - in the unique case $(x,y)=(0,0)$". $\endgroup$ – Oleg567 Apr 3 '13 at 10:07
  • 1
    $\begingroup$ How did you get that solution? What transformations (edits) are needed? $\endgroup$ – Buksy Apr 3 '13 at 11:38
  • 1
    $\begingroup$ @Buksy: 1) $(x \pm y)^2 = x^2 \pm 2xy + y^2$. 2) $x^2+xy+y^2$ is ~ like 1), but with multiplier 1 instead of $\pm 2$. 3) $4(x^2+xy+y^2) = \alpha (x+y)^2 + \beta (x-y)^2$. 4) to fit $\alpha$, $\beta$. $\endgroup$ – Oleg567 Apr 3 '13 at 11:49
16
$\begingroup$

Why not simply $x^2+xy+y^2=(x+\frac{y}{2})^2+\frac{3}{4}y^2>0$?

$\endgroup$
  • $\begingroup$ Yes, it is similar too. Nice observation. In general (if we talk about real values), our formulas can be described as: let $a$ $-$ any angle, $b=a\pm\pi/3$. So, $(x\cos{a}+y\cos{b})^2+(x\sin{a}+y\sin{b})^2$ = $(\cos^2{a}+\sin^2{a})x^2 + 2(\cos{a}\cos{b}+\sin{a}\sin{b})xy+(\cos^2{b}+\sin^2{b})y^2$ = $x^2+2\cos(\pm \pi/3)xy+y^2 = x^2+xy+y^2$. Your case: $a=0, b=\pi/3$ $-$ more short (has zero). My case: $a=\pi/6, b=-\pi/6$ $-$ more symmetric. $\endgroup$ – Oleg567 Apr 4 '13 at 9:35
  • $\begingroup$ @Oleg567, I think you just want $x=x'\cos\alpha+y'\sin\alpha,y=x'(-\sin\alpha)+y'\cos\alpha$, which is geometrically understood as a rotation about the origin. But what I did was just simply completing the squares, this is a very useful trick for treating quadratics. $\endgroup$ – Easy Apr 5 '13 at 9:21
13
$\begingroup$

As you say, if $x,y$ have the same sign then: $x^2+xy+y^2=(x-y)^2+3xy>0$. If $x,y$ have opposite signs then $x^2+xy+y^2=(x+y)^2-xy>0$.

$\endgroup$
  • $\begingroup$ Thanks a lot ,sir.Great concept. $\endgroup$ – learner Apr 3 '13 at 9:38
9
$\begingroup$

Hint: $\displaystyle x^2+xy+y^2=y^2 \left( \left( \frac{x}{y} \right)^2+ \frac{x}{y} +1 \right)$; so you just have to study the polynomial $P(z)=z^2+z+1$.

$\endgroup$
  • $\begingroup$ I can't see what this is getting at. $\endgroup$ – Hammerite Apr 3 '13 at 10:31
  • $\begingroup$ @Hammerite: You have $x^2+xy+y^2=y^2P(y/x)$, with $P(z)=z^2+z+1$; but $P(z)>0$ (since its discriminant is negative). $\endgroup$ – Seirios Apr 3 '13 at 11:06
  • $\begingroup$ I see. Could you make a minor edit to your answer, this site wo't let me cancel my downvote unless you do. $\endgroup$ – Hammerite Apr 4 '13 at 12:34
  • $\begingroup$ @Hammerite: It's done. $\endgroup$ – Seirios Apr 4 '13 at 13:00
5
$\begingroup$

I am posting 2 ways of solving this:

$(1)$ If $x,y$ belong to positive real numbers only:

We know that $x^2 + y^2 \geq 2xy$.

Hence we can say that $x^2 + y^2 > xy$

Hence even if $xy$ is negative $x^2 + y^2$,which is positive, is always greater than $xy$ making the sum $x^2 + y^2 + xy$ always positive.

$(2)$ If $x,y$ belong to real numbers:

$$x^2 + y^2 + xy$$

$$ = x^2 + 2(x)(\dfrac{y}{2}) + \dfrac{y^2}{4} + \dfrac{3y^2}{4}$$

$$={(x+\dfrac{y}{2})}^2 + \dfrac{3y^2}{4}$$which is always positive.

Hope the answer is clear now !!

$\endgroup$
  • $\begingroup$ @learner: Please do accept this (or any better) solution so that the question is marked as solved. $\endgroup$ – user65119 Apr 3 '13 at 9:43
  • 1
    $\begingroup$ (-1) as, say, $xy = -3$ and $x^2 + y^2 = 1$ is not excluded by your reasoning (but in which case $x^2 + xy + y^2 < 0$) $\endgroup$ – TMM Apr 3 '13 at 10:18
  • 1
    $\begingroup$ Please note $x^2+y^2 \ge 2xy$ does not mean for real numbers $x,y$ $x^2+y^2>xy$ as $2xy>xy $ is valid only when $xy>0$ or both $x,y$ are of same sign. So your proof is valid only when both $x$ and $y$ are of the same sign. $\endgroup$ – jdoicj Apr 3 '13 at 12:56
  • 1
    $\begingroup$ TMM and boywholived are both right -- this is not a valid solution. $\endgroup$ – Jonathan Apr 3 '13 at 14:10
  • $\begingroup$ @lsp you can look at math.stackexchange.com/a/349948/63095 which is the solution posted by "Ittay Weiss" and look how he got rid of the case when $x$ and $y$ both real numbers of opposite sign. $\endgroup$ – jdoicj Apr 3 '13 at 15:33
4
$\begingroup$

Don't give up so soon! Your idea also works when they have opposite sign:

$$ x^2+xy+y^2=(x+y)^2-xy>0$$

$\endgroup$
  • $\begingroup$ This really is the most natural answer by far (for this particular OP). It also captures, more easily than the diagonalization answers, the range of $c$ for which $x^2+cxy+y^2$ is positive definite. $\endgroup$ – Erick Wong Apr 4 '13 at 7:29
3
$\begingroup$

$x^2 + xy + y^2$ is a quadratic form and can be written $$x^2 + xy + y^2 = \begin{bmatrix} x & y\end{bmatrix}\begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}$$

A matrix $\mathbf{A}$ is positive-definite if $\mathbf{z'}\mathbf{A}\mathbf{z}>0 $ for all $\mathbf{z}\neq 0$

Show that $\begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{bmatrix}$ is a positive-definite matrix and you will have a very nice solution to your problem

$\endgroup$
  • $\begingroup$ Thanks a lot,Gabra. Got it. +1 from me.Nice approach. $\endgroup$ – learner Apr 3 '13 at 12:54
  • $\begingroup$ By making an eigen-decomposition of A above you will get the solution provided by @Oleg567. $\endgroup$ – Gabra Apr 3 '13 at 20:22
2
$\begingroup$

Some of the answers here are rigorous but overly complicated. Consider that $x^2$ is positive and $y^2$ is positive. The only term that could be negative is $xy$.

Suppose that $xy$ is negative. If abs($x$)>abs($y$), then $x²>-xy,$ so the whole expression is positive. The same logic applies for abs($y$)>abs($x$). The $x=y$ case is clearly positive.

Therefore the whole expression is positive.

$\endgroup$
0
$\begingroup$

Writing the given expression in the quadratic matrix form: $$ x^2 + xy + y^2 = \begin{bmatrix} x \\ y \end{bmatrix}^T \underbrace{\begin{bmatrix} 1 & a \\ (1-a) & 1 \end{bmatrix}}_{A} \begin{bmatrix} x \\ y \end{bmatrix} \quad\quad\quad (a \in [0,1]) $$

According to the Gershgorin Circle Theorem, both the eigenvalues of $A$ may lie in $[0,2]$, and two of them cannot be zero at the same time; so A is at least positive semi-definite.

But, if we manually check for extreme cases (i.e.; $a=0$ and $a=1$) for a more strict analysis, we see that both the eigenvalues are $1$ when $a=1$ and $a=0$.

Therefore, $A$ is positive definite. The given expression is always positive under the given conditions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.