Prove x^4+x^3y+x^2y^2+xy^3+y^4 > 0 when x,y aren't both zero

[silina01]

Prove that if x and y are not both , then x^4+x^3y+x^2y^2+xy^3+y^4 > 0

I have no idea how to start this proof, can anyone give me an idea?

 

[Dick]

Here's a hint for one way to do it. If x=0 it's clearly true. If x is not zero, then factor x^4 out and look at the other factor. Remind you of anything?

 

[silina01]

sorry I meant to say that if x and y are both not 0

 

[dirk_mec1]

Hint: This expression can be simplified.

 

[silina01]

I factored out x^4 and got x^4 [(y/x)^4 + (y/x)^3 +(y/x)^2 + y/x +1] I see that all these terms have y/x so if i let t = y/x then it'll be x^4 [t^4 + t^3 +t^2 + t +1] (just so it is easier to visualize) but I am stuck, How do i simplify it further?

 

[dirk_mec1]

Hint: think about a certain sequence.

 

[silina01]

I am still lost

 

[dirk_mec1]

What special sequences do you know of?

 

[silina01]

x^n - y^n = (x - y) (x^n-1 + x^n-2y+ ...+ xy^n-2 + y^n-1 )??

 

[haruspex]

x^n - y^n = (x - y) (x^n-1 + x^n-2y+ ...+ xy^n-2 + y^n-1 )??

That'll do it. So how can you use that to rewrite the original expression?

 

[silina01]

That'll do it. So how can you use that to rewrite the original expression?

(x^5 - y^5)/(x - y) = x^4 + x^3y + x^2y^2 + xy^3 + y^4 , is that correct?

 

[silina01]

ahhhh I see it know, in both cases if x-y <0 or if x-y>0 the quotient will always be positive. Thanks everyone.

 

[haruspex]

ahhhh I see it know, in both cases if x-y <0 or if x-y>0 the quotient will always be positive. Thanks everyone.

Right, but there is one special case you need to address separately.

 

[silina01]

what would that be?

 

[silina01]

x and y cannot be 0 by the way

 

[haruspex]

what would that be?

Your expression involves 1/(x-y). What doubt should that create?

 

[Dick]

x and y cannot be 0 by the way

But they can be nonzero and equal.

 

[dirk_mec1]

But they can be nonzero and equal.

But if you simplify the fraction then there is no problem.

 

[Dick]

But if you simplify the fraction then there is no problem.

I didn't say there was a problem, but x=y is a special case.

 

[Gzyousikai]

Assume that $$x,y>0$$.Let $$P(x,y)$$be the polynomial, then $$P(x,y)=P(-x,-y)\ge P(x,-y)=x^4+y^4-x^3y-xy^3+x^2y^2\ge x^2y^2\ge 0$$ And the equality cannot hold.