# Thread: Prove function is decreasing

1. ## Prove function is decreasing

Hi, I'm trying to prove that $f(x) = \frac{e^x}{x^x}$ is decreasing on interval $[1, \infty)$.

Know that the statement I want to prove is as follows: $\forall a, b \in [1, \infty), a < b \Rightarrow \frac{e^a}{a^a} \geq \frac{e^b}{b^b}$.

I start off with a < b, then try to reach $\frac{e^a}{a^a} \geq \frac{e^b}{b^b}$ through some algebraic manipulations.

$$a < b$$
$$a^a < b ^b$$
$$\frac{1}{a^a} > \frac{1}{b^b}$$
$$\frac{e^a}{a^a} > \frac{e^a}{b^b}$$
Then what?

2. ## Re: Prove function is decreasing

$x^x = e^{\ln(x) x}$

$\dfrac{e^x}{x^x} = \dfrac{e^x}{e^{x \ln(x)}} = e^{x(1-\ln(x))}$

Showing this is decreasing is pretty trivial. See what you can manage.

3. ## Re: Prove function is decreasing

Originally Posted by romsek
$x^x = e^{\ln(x) x}$

$\dfrac{e^x}{x^x} = \dfrac{e^x}{e^{x \ln(x)}} = e^{x(1-\ln(x))}$

Showing this is decreasing is pretty trivial. See what you can manage.
Thanks a lot for your reply. This is what I have so far...

Assume $a < b$.
We want to show $e^{a(1-ln(a))} \geq e^{b(1-ln(b))}$
We know $1 - ln(a) > 1 - ln(b)$
Then $b(1 - ln(a)) > b(1 - ln(b))$

Then what? I am facing a similar problem as I had before. I can't say $a(1 - ln(a))> b(1 - ln(a))$, even though I would like it to be.

4. ## Re: Prove function is decreasing

Look at the first derivative of

$f(x) = e^{x(1-\ln(x))}$
$f^\prime(x) =-\ln(x)e^{x(1-\ln(x))} < 0,~\forall x > 1$

are you not allowed to use calculus?

5. ## Re: Prove function is decreasing

Ah, good point. Thanks