Prove the basic algebraic inequality used extensively everywhere.

Question

How can I prove rigourously that if for any real numbers $x$ and $y$ (both are positive)

If $x \ge y$ then

This implies $x^n \ge y^n$ for any real $n$ as well. (n is not negative) This is used everywhere for example when n is half or two.

Is there a general result if $n$ is negative?

Answer 1

What if $y<x<-1 \cup n\mod 2 \equiv 0$? In that case, $x^n<y^n$, disproving the function.

Answer 2

First of all, this inequality is only true for $n>0$. If $n$ is negative then exponentiation is order-reversing.

Second, an outline of the proof goes something like this:

1. First, prove the result for whole number values of $n$. This can be done by induction, making use of the fact that multiplication by a positive number is order-preserving.
2. Next, use the result above to prove the inequality for $n$ of the form $1/m$, with $m$ a whole number.
3. Combine the two results to prove it is true for all positive rational exponents.
4. Use continuity to extend the definition to all positive real exponents.

Answer 3

Taking advantage of the strict monotonicity of $f(x)=\ln{x}$ then for any $n>0$ and for any $x,y>0$ we have WLOG $$ x\ge y \iff \ln{x}\ge\ln{y} \iff n\ln{x}\ge n\ln{y} \iff \ln{x^n}\ge\ln{y^n} \iff x^n \ge y^n. $$