If $x$ and $y$ are not both $0$ then $ x^2 +xy +y^2> 0$

Question

Can't quite finish this proof:

Prove that if $x$ and $y$ are not both $0$ then $ x^2 +xy +y^2> 0$

$ x^2 +xy +y^2 +xy -xy> 0$

$ (x +y)^2 -xy> 0$ Without loss of generality define $x\geq y$

Firstly the case where $y<0$ and $x>0$ is trivial as both $ (x +y)^2$ and $-(xy) >0$.

I want to say by transitivity as each are >0 that there sum must be ${}> 0$ can I do that?

The second trivial case is $y=0$ as we simply have $ x^2 >0 $ where $x>0$.

Lastly we have define $c= x+y$ since both $x$ and $y$ must have the same sign we know $|c|>x$ since $x\geq y$ we know that $ x^2 \geq xy$ finally since $ |c|^2 = c^2$ $ \forall c \in \mathbb{R} $

We have $ c^2 >x^2 \geq xy$ hence $c^2 -xy >0$

I feel like I have all the pieces but it doesn't feel finished how do I fix it?

Edit: I am really glad i asked this question love the variety of different good answers no idea what a discriminant is but im going to go look that up. I wanted to give the accepted answer to Edward as he actually answered my question but Anurag s proof was just so much better then my attempt. ^^

One way is $x^2+xy+y^2 = \frac{3}{4}(x+y)^2 + \frac{1}{4}(x-y)^2 > 0$ of not both $x,y$ are $0$. Another way is looking at the discriminant of the quadratic. — r9m, Dec 19 '14 at 02:58
Either $x^2 \ge |x||y|$ or $y^2 \ge |x||y|$, so $x^2 + xy + y^2 \ge x^2 - |x||y| + y^2 > 0$. — MackTuesday, Dec 19 '14 at 08:49
Almost the same question: http://math.stackexchange.com/questions/920605/inequality-x2y2xy-ge-0 — Chris Culter, Dec 19 '14 at 09:31

score 19 · Accepted Answer · answered Dec 19 '14 at 02:59

19

Consider $T=x^2+xy+y^2$. then $2T=x^2+y^2+(x+y)^2 \geq 0$. The only way $T=0$ is when $x=0,y=0$, but that has already been ruled out by the conditions given.

answered Dec 19 '14 at 02:59

Anurag A

41,067

score 8 · Answer 2 · answered Dec 19 '14 at 03:25

8

Here's another way:

If $x=y$, then obvious. Otherwise: $$x^2+xy+y^2=\frac{x^3-y^3}{x-y}>0$$ as $f(x)=x^3$ is monotonically increasing.

answered Dec 19 '14 at 03:25

Saibal

1,597

score 6 · Answer 3 · answered Dec 19 '14 at 02:59

6

You can simply note that $$ x^2+xy+y^2=(x+y/2)^2+3y^2/4\geq 0. $$ Equality is iff $x=y=0$. So if $x,y$ are not both $0$, then $x^2+xy+y^2>0$.

answered Dec 19 '14 at 02:59

Kim Jong Un

14,794
1
24
49

score 4 · Answer 4 · answered Dec 19 '14 at 02:58

4

Let $y$ a fixed non zero real. The discriminant of the quadratic expression $x^2+xy+y^2$ is $\Delta=y^2-4y^2=-3y^2< 0$ hence the expression doesn't change the sign and we have $$x^2+xy+y^2>0$$

answered Dec 19 '14 at 02:58

score 2 · Answer 5 · answered Dec 19 '14 at 02:59

2

Your proof is correct. The first two cases are obvious. The third case follows directly from the AM-GM inequality:

$$\frac{x+y}{2}\ge \sqrt{xy}$$ $$\Rightarrow x+y>\sqrt{xy}$$

answered Dec 19 '14 at 02:59

Edward Jiang

3,670

Whats the full name of this inequality so i can look it up? – Faust Dec 19 '14 at 03:12
@Faust7: Inequality of arithmetic and geometric means – Dennis Dec 19 '14 at 03:51

score 2 · Answer 6 · answered Dec 19 '14 at 03:09

2

We only need to consider 2 cases:-

(1) Both x and y are negative

Then, $x^2 + xy + y^2 = (x – y)^2 + 3xy = (+ve) + (+ve) > 0$

(2) Only one is negative

Then, $x^2 + xy + y^2 = (x + y)^2 - xy = (+ve) – (-ve) > 0$

answered Dec 19 '14 at 03:09

Mick

17,141

what's (+ve) stand for/mean? – Faust Dec 19 '14 at 03:10
@Faust7 A short form to mean the value of a quantity is positive. – Mick Dec 19 '14 at 03:12

score 2 · Answer 7 · answered Dec 19 '14 at 09:32

2

Switching to polar coordinates ($x=r\cos \theta,y=r \sin \theta)$, we need to prove $$1+\frac{1}{2} \sin 2 \theta>0, $$ which is obvious.

answered Dec 19 '14 at 09:32

user1337

24,381

If $x$ and $y$ are not both $0$ then $ x^2 +xy +y^2> 0$

7 Answers7

Linked

Related