If $x \neq 0,y \neq 0,$ then $x^2+xy+y^2$ is .....

Question

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.

You can check the sign of discriminant and conclude . – hrkrshnn Apr 03 '13 at 12:54 — hrkrshnn, Apr 03 '13 at 12:54

score 37 · Accepted Answer · answered Apr 03 '13 at 09:36

37

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

answered Apr 03 '13 at 09:36

Oleg567

17,295

2

Lovely solution! (+5 if I could) – Ittay Weiss Apr 03 '13 at 09:38
Wow! got it.Thanks a lot. +1 from me. – learner Apr 03 '13 at 09:40
3

Why did you say "when $x\neq0$ and $y\neq0$"? You need neither for the equality, and only one of them for the inequality. – Marc van Leeuwen Apr 03 '13 at 09:54
1

@MarcvanLeeuwen, yes, answer can be expanded with words "equality - in the unique case $(x,y)=(0,0)$". – Oleg567 Apr 03 '13 at 10:07
1

How did you get that solution? What transformations (edits) are needed? – Buksy Apr 03 '13 at 11:38
1

@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 $\alpha$, $\beta$. – Oleg567 Apr 03 '13 at 11:49
1

@MarcvanLeeuwen: Because the problem statement specified that neither are $0$. – Joe Z. Apr 03 '13 at 13:05
@learner Note that this is no longer true if complex numbers are allowed, in which case any value is possible, be it negative, zero or complex valued again - just solve $x^2+xy+y^2=c$ for $x$ to obtain $2x_{1,2} = -y\pm \sqrt{4c-3y^2}$. That's also a way to see that for $c=0$ there are no real valued $x,y\neq0$ possible, which means the sign (for real $x,y\neq0$) is always the same and you can just test one pair, e.g. (1,1). – Tobias Kienzler Apr 04 '13 at 06:49

score 16 · Answer 2 · answered Apr 03 '13 at 09:52

16

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

answered Apr 03 '13 at 09:52

Easy

4,485

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.
– Oleg567 Apr 04 '13 at 09:35
@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. – Easy Apr 05 '13 at 09:21

score 13 · Answer 3 · answered Apr 03 '13 at 09:35

13

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

answered Apr 03 '13 at 09:35

Ittay Weiss

79,840
7
141
236

Thanks a lot ,sir.Great concept. – learner Apr 03 '13 at 09:38

Seirios · Answer 4 · 2013-04-04T13:00:47.817

9

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 $P(z)=z^2+z+1$.

edited Apr 04 '13 at 13:00

answered Apr 03 '13 at 09:30

Seirios

33,157

I can't see what this is getting at. – Hammerite Apr 03 '13 at 10:31
@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). – Seirios Apr 03 '13 at 11:06
I see. Could you make a minor edit to your answer, this site wo't let me cancel my downvote unless you do. – Hammerite Apr 04 '13 at 12:34
@Hammerite: It's done. – Seirios Apr 04 '13 at 13:00

score 5 · Answer 5 · answered Apr 04 '13 at 06:29

5

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

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

answered Apr 04 '13 at 06:29

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. – Erick Wong Apr 04 '13 at 07:29

lsp · Answer 6 · 2013-04-04T06:23:06.783

4

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 !!

edited Apr 04 '13 at 06:23

answered Apr 03 '13 at 09:32

lsp

4,745

@learner: Please do accept this (or any better) solution so that the question is marked as solved. – Apr 03 '13 at 09:43
1

(-1) as, say, $xy = -3$ and $x^2 + y^2 = 1$ is not excluded by your reasoning (but in which case $x^2 + xy + y^2 < 0$) – TMM Apr 03 '13 at 10:18
1

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. – hrkrshnn Apr 03 '13 at 12:56
1

TMM and boywholived are both right -- this is not a valid solution. – Jonathan Apr 03 '13 at 14:10
@lsp you can look at http://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. – hrkrshnn Apr 03 '13 at 15:33

score 3 · Answer 7 · answered Apr 03 '13 at 12:51

3

$x^2 + xy + y^2$ 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

answered Apr 03 '13 at 12:51

Gabra

131

Thanks a lot,Gabra. Got it. +1 from me.Nice approach. – learner Apr 03 '13 at 12:54
By making an eigen-decomposition of A above you will get the solution provided by @Oleg567. – Gabra Apr 03 '13 at 20:22

score 2 · Answer 8 · edited Apr 03 '13 at 14:03

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.

score 0 · Answer 9 · answered Apr 04 '13 at 06:42

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.

If $x \neq 0,y \neq 0,$ then $x^2+xy+y^2$ is .....

9 Answers9

Linked