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Let $f(x)=e^{-x} \sqrt{x}$ , $g(x)=\frac{x^2-2ax+1}{x+1}$ and $a>\ln 2-1$. Prove $f(x)g(x)<\frac{1}{2}$.

This problem is one of my homework. After my studying, I know $f(x)_{max}=\frac{\sqrt{2e}}{2e}<\frac{1}{2}$.

C. Simon
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1 Answers1

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I'm not sure this is the easiest way to solve this, but here are my 2 cents. I'm assuming x>0, since x=0 results in dividing by 0 and x<0 gives complex numbers, and i'm thinking you're are working with $ \mathbb{R} $

$$ J(x) := f(x) \cdot g(x) = e^{-x} \sqrt{x} \frac{x^2 - 2ax+1}{x+1} $$ Looking at the function it's clear in order to find the max, you want to set a as low as possible. So setting a= ln(2)-1. (I know, that is not possible, but if i find the max of this to be lower than 0.5, it must be true for $\forall a $. $$ J(x) := e^{-x} \sqrt{x} \frac{x^2 - 2 ( ln(2)-1)x+1}{x+1} $$ I want to find $ J(x)_{max} $

So first want to find critical points, and that is to be honest a very complicated calculation.

$ \frac{d}{dx}J(x) = 0 $, x=1 . hope someone have a better way to find this, because i relied a lot on my calculator.

Now you only have to check end point and critical points. $$ J(1) = 0.5 e^{-1} (4-ln(2) < 0.5 $$ $$ \lim\limits_{x \rightarrow 0 +} J(x) =0 $$ $$ \lim\limits_{x \rightarrow \infty} J(x) =0 $$ Hope this helps

Viktor Jeppesen
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    Thank you for your help.

    I calculated that $$\frac{d}{dx}J(x)=\frac{\sqrt{x}(x-1)e^{-x}}{2x(x+1)}\big(-2x^3-(x-1^2)+(\ln2-1)(4x^2+6x)\big).$$ Therefore, if $0<x<1$, then $\frac{d}{dx}J(x)>0$. And, if $x>1$, then $\frac{d}{dx}J(x)<0$. From this result, we have $$J(x)_{max}=J(1)=\frac{2-\ln 2}{e}<\frac{1}{2}.$$

     – C. Simon Jun 02 '16 at 07:54
  • Seems like a good way to come around the very complicated calculation. Remember 2 things.
    1. You are not sure, thats J(1) is your max point before checking J''(x) or checking the values in end points.
    2. In order to say, that the critical point is x=1 (with the calculation arguments you just gave), you need J'(x) to be continuous. You can just give a simple argument, why this is true.
     – Viktor Jeppesen Jun 02 '16 at 11:44