# Thread: Exponential Inequality: Help me!!

1. ## Exponential Inequality: Help me!!

Prove that for all positive real numbers the following inequality holds:

2. if: $\displaystyle x > y$ then $\displaystyle x^x>y^x$ and $\displaystyle x^y>y^y$ and it can be deduced that $\displaystyle x^x-y^x>x^y-y^y$

So if we change things around we get: $\displaystyle x^x-y^x>x^y-y^y\quad\rightarrow\quad x^x+y^y>x^y+y^x$

So that's true for $\displaystyle x>y$

if: $\displaystyle x=y$ then $\displaystyle x^x=x^y$ and $\displaystyle y^y=y^x$ so then $\displaystyle x^x +y^y=x^y+y^x$

So that's true for $\displaystyle x=y$

if: $\displaystyle x <y$ then $\displaystyle x^x<y^x$ and $\displaystyle x^y<y^y$ and it can be deduced that $\displaystyle y^y-x^y>y^x-x^x$

So if we change things around we get: $\displaystyle y^y-x^y>y^x-x^x\quad\rightarrow\quad y^y+x^x>y^x+x^y$

So it holds true in all cases.

3. Hello Quick,
Originally Posted by Quick
if: $\displaystyle x > y$ then $\displaystyle x^x>y^x$ and $\displaystyle x^y>y^y$ and it can be deduced that $\displaystyle x^x-y^x>x^y-y^y$
Huh ? How can this be deduced ?

You know that $\displaystyle x^x-y^x>0$ and $\displaystyle x^y-y^y>0$.

Ok.. But if $\displaystyle a>0$ and $\displaystyle b>0$, does $\displaystyle a>b$ ???

4. Hello, Moo.

Originally Posted by Moo
Hello Quick,

Huh ? How can this be deduced ?

You know that $\displaystyle x^x-y^x>0$ and $\displaystyle x^y-y^y>0$.

Ok.. But if $\displaystyle a>0$ and $\displaystyle b>0$, does $\displaystyle a>b$ ???
let's say that: $\displaystyle a >b$

so let's take a look at: $\displaystyle a^m-b^m$

It can be expanded into: $\displaystyle (a-b)(a^{m-1}+a^{m-2}b+...+ab^{m-2}+b^{m-1})$

so let's take a look at: $\displaystyle a^n-b^n$

It can be expanded into: $\displaystyle (a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})$

and let's say that: $\displaystyle m>n$

alright, now we want to know if the following is true: $\displaystyle a^m-b^m>a^n-b^n$

Well if we expand it: $\displaystyle (a-b)(a^{m-1}+a^{m-2}b+...+ab^{m-2}+b^{m-1})>(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})$

Then divide away (a-b) to get: $\displaystyle a^{m-1}+a^{m-2}b+...+ab^{m-2}+b^{m-1}>a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1}$

and that expression is clearly true.

Therefore, it can be deduced that if $\displaystyle x>y$ then $\displaystyle x^x-y^x>x^y-y^y$

I understood that I had left that step rather vague before you posted. I was hoping that either the questioner would understand or someone else would pick it up and explain it in more detail.

It is impolite to make the assumption that there is no logic behind my statement, and it would have been best to either ask for my explanation or attempt a proof yourself.

5. Thank You so much!!! Nice and understandable proof!!!