Prove that $(2-x)^nx^{n-1}$ decreases with $n$ for $0 How can I show that:
$$(2-x)^nx^{n-1}$$ is decreasing with $n$ when $0<x<1$?  I think this is generally true, but specifically I am concerned with $n$ as an integer $\geq 2$ and showing that the maximum of the function is when $n=2$ (its minimum) for all $x$.
When I take the derivative with respect to $n$, I just get  $$(2 - x)^n x^n \log(2 - x) + (2 - x)^n x^n \log x ,$$ but I don't know how to show that that is negative either. 
I guess it comes down to showing that the absolute value of $\log(2-x)$ is less than the absolute value of $\log(x)$... but I don't know how to do that with logs, or if that's necessarily the right approach.
 A: Your idea works, you just have to push it a little farther. Take the derivative with respect to $n$, then consolidate like terms and put the two logarithms together for
$$(2-x)^nx^n \log ((2-x)x).$$
The factor $(2-x)^n x^n$ is positive so it remains to be seen that $0<(2-x)x<1$ for $x\in(0,1)$. We can subtract one and factor to get $-1<-(x-1)^2<0$, which is obviously true in our case.
A: $$(2-x)^nx^{n-1} = \frac1x \Big(x(2-x)\Big)^n.$$
Everything in parentheses here is positive.  If you can show that the expression raised to the power $n$ is between $0$ and $1$, you're done.  $y=x(2-x)$ is a parabola opening downward with $x$-intercepts at $0$ and $2$, and parabolas are symmetric, so the vertex is half-way between $0$ and $2$.  That's the highest point.  When $x=1$, then $y=1$.  So $y<1$ if $x=\text{anything else}$.
A: Is $n$ a whole number or a real?  Usually it would be a whole number.  If so, to show it is decreasing with $n$, you need to show that the multiplicative factor, $(2-x)x$ is less than $1$.  This, with the fact that the basic term is greater than $0$, is enough.
If $n$ is real, you just need to show that $\log x +\log(2-x) \lt 0$  as the other terms are positive and distribute out.  You can check this with a derivative test.
