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

[nietzsche]

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

Homework Statement

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

Homework Equations

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]

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]

(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?

 

[nietzsche]

Okay, here's another attempt.

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

 

[nietzsche]

(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?

 

[aPhilosopher]

a worse idea for a proof used to live here.

 

[nietzsche]

Thank you all!

 

[VietDao29]

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. :)

 

[njama]

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

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

 

[emyt]

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

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

could you show your work please?

thanks

 

[njama]

could you show your work please?

thanks

I've already showed it.

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

 

[emyt]

I've already showed it.

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)

 

[nietzsche]

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