Inequality X^n<Y^n if x<y and n is odd

1. Mar 16, 2015

Karim Habashy

Hi All,

Question : Prove that xn<yn , given that x<y and n is odd .

Attempt at solution :
Assumptions:
y-x>0
y2>0
x2>0

So y2x<y3 & x3<x2y

So i need to prove that x2y<y2x

i.e need to prove

then y2x-x2y>0

then yx(y-x)>0, from assumptions y-x>0 so i need to prove that yx>0, so i have 3 cases

Case 1 : y and x are both positive so the inequality is true and everything is true.
Case 2: y and x are both negative so the inequality is true again and everything is true
Case 3 : y is positive and x is negative (as y>x), then the above inequality is not true, but i go back knowing that y is positive and x is negative, then y>0 and x<0 so y3>0 and x3<0 so y3>x3

At the end x3<y3 and by repetition xn <yn

End.

Is this proof true ? is there an easier way.

Thanks.

2. Mar 16, 2015

Staff: Mentor

Looks fine.

3. Mar 17, 2015

mathman

There may be an easier way, by considering 3 cases.
1. both positive. $y^n-x^n=(y-x)(x^{n-1}+yx^{n-2}+...y^{n-1}) \gt 0$ Since y-x > 0 and series has all positive terms.
2. y >0 and x < 0. Odd powers have same sign.
3. both negative. Change sign for both and use case 1 to show $-x^n \gt -y^n$.

4. Mar 18, 2015

da_nang

Alternatively, show that f(x) = x2k+1 for all natural numbers k is strictly monotonically increasing.

Last edited: Mar 18, 2015