Is there a way to prove this exponential inequality: if $a>b$ then $a^a>b^b$ for $a,b>1$?

Question

I came across this proposition while trying to prove that a function was injective: if $a>b$ then $a^a>b^b$, where $a$ and $b$ are real numbers bigger than $1$. Intuitively it (somehow) makes sense but I wonder if a rigorous proof can be made. But, the initial problem I was trying to solve was to show that $f(x)=x^x$, where $x$ is just a positive real number, is injective. As the "contrapositive method" from the definition of an injective function didn't work out, I figured I could just show that my function was strictly increasing or decreasing, therefore the function would be injective. I looked at the graph of this function and I noticed I have a turning point at $x=1/e$ (as the user MXYMXY pointed out). Thus I had two cases for my function.

Answer 1

If $a\gt b$ and $a\gt1$ and $b\gt1$ and $a,b\in\mathbb{R}$ then this implies:$$a^b\gt b^b\tag{1}$$

Also, if $a\gt b$ then this implies:$$a^a\gt a^b\tag{2}$$

Now, making use of result (1) in result (2) yields:$$a^a\gt a^b\gt b^b$$

Answer 2

Try this sequence of inequalities: $$b^b < b^a < a^a $$

Answer 3

Note that the derivative of $x^x$ is $(\ln(x)+1)x^x>0$(see here) if $x>e^{-1}$, implying $x^x$ is a strictly increasing function if $x>e^{-1}$.

This implies your claim is true not only when $a,b>1$ but also $a,b>e^{-1}$

Answer 4

You're asking to prove that the map $f(x):=x^x=\exp(x\log x)$ is strictly increasing. But since $x\mapsto x$ and $x\mapsto\log x$ are both strictly increasing and the first one is $>1$ in $\Bbb R_{>1}$, their product is such (in $\Bbb R_{>1}$).

Then the exponential is strictly increasing too, so $f$ is the composition of strictly increasing maps, thus itself is strictly increasing too.

Answer 5

Since $a > b > 1$, and since $\log(x)$ is an increasing function of $x$, then $$\log(a) > \log(b) > 0.$$ Multiplying these two inequalities together: $$a > b > 1$$ and $$\log(a) > \log(b) > 0$$ gives you $$a\log(a) > b\log(b).$$ By a property of logarithms ($x\log(x) = \log(x^x)$), this implies $$\log(a^a) > \log(b^b).$$ Again, since $\log(x)$ is an increasing function of $x$, we obtain $$a^a > b^b.$$

QED