Another proof: x^2 + xy +y^2 > 0

Hello again, I have another proof that I can't figure out how to solve.

Homework Statement

$$\text{Prove that if }\textit{x }\text{and }\textit{y }\text{are not both 0, then} \begin{equation*} x^2+xy+y^2>0\tag{1} \end{equation*}$$

N/A

The Attempt at a Solution

Not sure if I'm on the right track here...

\text{Assume (1) is true:} \begin{align*} x^2+xy+y^2 &> 0\\ x^2 +2xy + y^2 &> xy\\ (x+y)^2 &> xy\\ \end{align*}

...not sure where to go from here.

Dick
Homework Helper
I would say to look for the extrema of f(x,y)=x^2+xy+y^2. Find df/dx and df/dy and set them equal to zero and solve for x and y. Can you classify that as a min or a max or neither?

Office_Shredder
Staff Emeritus
Gold Member
(x+y)2 = x2 + 2xy + y2 >= 0 You know that already

So

x2 + xy + y2 >= -xy

If x and y are both positive, the result is trivial. If x and y are both negative, the result is also trivial. (in both cases, each term in the summation is positive). When one of x or y is negative, -xy becomes positive. So what can you say?

Okay, here's another attempt.

\begin{align*} x^3 - y^3 &= (x-y)(x^2+xy+y^2)\\ \frac{x^3 - y^3}{x-y} &= x^2+xy+y^2 \tag{1}\\ \end{align*} \begin{align*} x^2 + xy + y^2 &> 0\\ \frac{x^3 - y^3}{x-y} &> 0 \tag{2} \end{align*} \text{(2) is true for } \begin{math}x<y\end{math} \text{, } \begin{math}x>y\end{math} \text{,} \begin{math}x, y \in \mathbb{R}.\end{math}\\ \\ \text{For } \begin{math}x=y\end{math} \text{:} \begin{align*} x^2+xy+y^2 &> 0\\ x^2 + x^2 + x^2 &> 0\\ 3x^2 &> 0\tag{3} \end{align*} \text{(3)} is true for all \begin{math}x, y \in \mathbb{R}.\end{math}

(x+y)2 = x2 + 2xy + y2 >= 0 You know that already

So

x2 + xy + y2 >= -xy

If x and y are both positive, the result is trivial. If x and y are both negative, the result is also trivial. (in both cases, each term in the summation is positive). When one of x or y is negative, -xy becomes positive. So what can you say?

When one of x or y is negative, the LHS will be greater than a positive number and therefore greater than zero?

a worse idea for a proof used to live here.

Last edited:
Thank you all!

VietDao29
Homework Helper
Well, I think you guys are complicating things a little bit here.

We can use the idea of completing squares to solve this problem. It goes like this:

$$x ^ 2 + xy + y ^ 2 = \left[ x ^ 2 + 2 x \left( \frac{1}{2} y \right) + \left( \frac{1}{2}y \right) ^ 2 \right] + \frac{3}{4} y ^ 2$$

I'm almost spilling out the answer. Since the OP has already had his own solution, this is just another way to tackle the problem.

So, what left is to determine when the equation holds. :)

There is much easier way to do it:

$$x^2+xy+y^2=(x+y/2)^2-y^2/4+y^2=(x+y/2)^2+3/4(y^2)=(x+y/2)^2+(\frac{y\sqrt{3}}{2})^2$$

Now

$$(x+y/2)^2 + (\frac{y\sqrt{3}}{2})^2 \geq 0$$

1 person
There is much easier way to do it:

$$x^2+xy+y^2=(x+y/2)^2-y^2/4+y^2=(x+y/2)^2+3/4(y^2)=(x+y/2)^2+(\frac{y\sqrt{3}}{2})^2$$

Now

$$(x+y/2)^2 + (\frac{y\sqrt{3}}{2})^2 \geq 0$$

thanks

thanks

if a2 $\geq$ 0 and b2 $\geq$ 0 then a2+b2$\geq$ 0

if a2 $\geq$ 0 and b2 $\geq$ 0 then a2+b2$\geq$ 0

never mind, it was just from completing the square from above (I only skimmed through the thread sorry)

There is much easier way to do it:

$$x^2+xy+y^2=(x+y/2)^2-y^2/4+y^2=(x+y/2)^2+3/4(y^2)=(x+y/2)^2+(\frac{y\sqrt{3}}{2})^2$$

Now

$$(x+y/2)^2 + (\frac{y\sqrt{3}}{2})^2 \geq 0$$

applause

wow, that's so concise