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Thread: Prove function is decreasing

  1. #1
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    Question 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?
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  2. #2
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    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.
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  3. #3
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    Re: Prove function is decreasing

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

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

    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.
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  4. #4
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    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?
    Thanks from topsquark and mathKid42
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  5. #5
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    Re: Prove function is decreasing

    Ah, good point. Thanks
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