A question about function $f(x)=e^{-x} \sqrt{x}$ and $g(x)=\frac{x^2-2ax+1}{x+1}$. 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 $f(x)_{max}=\frac{\sqrt{2e}}{2e}<\frac{1}{2}$.
 A: 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
