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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.

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  • $\begingroup$ I didn't read the question properly at first. Sorry. $\endgroup$ – Foobaz John Feb 14 '16 at 0:44
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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:$$a^a\gt a^b\gt b^b$$

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    $\begingroup$ @BenjaminDickman read the restriction in the question, that a and b are both greater than 1. $\endgroup$ – Patrick Roberts Feb 14 '16 at 6:27
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    $\begingroup$ The notation $(a,b) \in \mathbb{R}$ is confusing, since $(a,b)$ looks as a pair. I'd write that without parentheses: $a,b \in \mathbb{R}$. $\endgroup$ – chi Feb 14 '16 at 18:40
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    $\begingroup$ @chi Thanks for heads up - duly noted and changed. $\endgroup$ – Mufasa Feb 14 '16 at 19:41
  • $\begingroup$ Why is assuming that $a^b > b^b$ and that $a^a > a^b$ any more "basic" than assuming that $a^a > b^b$, I wonder? $\endgroup$ – fgp Feb 14 '16 at 22:15
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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.

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    $\begingroup$ You should point out where you are using $x>1$. It is not true in general that the product of two strictly increasing functions is strictly increasing, take as an example the functions $x$ and $x$ with product $x^2$. $\endgroup$ – Carsten S Feb 14 '16 at 13:48
  • $\begingroup$ You're right, thanks. I'm gonna fix $\endgroup$ – Joe Feb 14 '16 at 14:02
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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 $a,b>e^{-1}$

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

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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

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