Solving for Prove That x^4+x^3y+x^2y^2+xy^3+y^4>0

[nietzsche]

Hi everyone. I know i keep posting all these questions, but each question in my textbook just keeps on bringing on new challenges. Is there a rule against posting to many questions?

Homework Statement

$$\text{Prove that if} \begin{math} x \end{math} \text{and} \begin{math} y \end{math} \text{are not both 0, then} \begin{equation*} x^4+x^3y+x^2y^2+xy^3+y^4>0 \end{equation*}$$

Homework Equations

N/A

The Attempt at a Solution

This is the second part of a question I posted earlier (https://www.physicsforums.com/showthread.php?t=338240). I'm guessing the questions are related somehow, but all the methods used on the earlier question don't seem to work on this question.

I tried grouping all the positive terms ($$x^4, y^4, x^2y^2$$) and separating the equation based on those, and it works for the cases where x and y are both positive or both negative, but when they have opposite signs, it's impossible to figure out.

Any hints? I know that multiplying by (x-y) gives $$x^5-y^5$$, but I don't know how much that helps.

 

[Elucidus]

Iis the last term in the inequality supposed to be y² or y⁴?

--Elucidus

 

[nietzsche]

Iis the last term in the inequality supposed to be y² or y⁴?

--Elucidus

ah yes, thank you, it's been fixed

 

[emyt]

it's just the same as the previous part isn't it?

 

[nietzsche]

it's just the same as the previous part isn't it?

well, I've tried this:

$$\begin{align*} x^4+x^3y+x^2y^2+xy^3+y^4 &> 0\\ x^4+y^4+xy(x^2+xy+y^2) &> 0 \end{align*}$$

and we know from the previous part that x^2 + xy + y^2 is positive, but there's that xy term which doesn't make it as clear cut as the last question.

 

[emyt]

(x^5 - y^5) / (x-y) ?

btw, check your inbox if you haven't

 

[Elucidus]

(x^5 - y^5) / (x-y) ?

Consider cases where x > y and x < y.

(The case where x = y is more easily dealt with in the original polynomial form.)

--Elucidus

 

[emyt]

yes, that's why I was confused - he came up with the same proof in his last thread

 

[nietzsche]

Consider cases where x > y and x < y.

(The case where x = y is more easily dealt with in the original polynomial form.)

--Elucidus

haha, i guess come up with that in my last thread. oops.