TITLE: Is there a way to prove this exponential inequality: if $a>b$ then $a^a>b^b$ for $a,b>1$?
QUESTION [10 upvotes]: 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.

REPLY [3 votes]: 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