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

[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.

 

[mfb]

Looks fine.

 

[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 $-x^n \gt -y^n$.

 

[da_nang]

Alternatively, show that f(x) = x^2k+1 for all natural numbers k is strictly monotonically increasing.