I have the following function,
$f(u)=1-a u^{a-1}(1-u)^b+b u^a(1-u)^{b-1} \text{for} \ u\in [0,1]\ \text{and} \ a,b\geq 1$,
and want to show $f(u)\geq0$ for $u\in [0,1]$.
It seems to be correct, as this is one ascertain from a paper and I have tried various numerous cases. Usual derivative technique does not work while I cannot find a nice inequality at this moment.
I put my way in the following. As you can see I failed in the last section.
\begin{align} g(u) &= 1 - a u^{a - 1} (1 - u)^b + b u^a (1 - u)^{b - 1}\\ &= 1 - u^{a - 1} (1 - u)^{b - 1} (a (1 - u) - b u)\\ &= 1 + u^{a - 1} (1 - u)^{b - 1} ((a + b) u - a). \end{align}
- If $u \in [\frac{a}{a + b}, 1]$, i.e., $(a + b) u - a \geq 0$ $\Rightarrow$ $u \geq \frac{a}{a + b}$, $g(u) \geq 0$.
- If $u = 0$, $g(u) = 1 \geq 0$.
- If $u \in [0, \frac{a}{a + b}]$. Consider $h(u) = u^{a - 1} (1 - u)^{b - 1} ((a + b) u - a)$.
3.1. Suppose $b \geq a$.
We first need a little lemma
For $0 \leq a \leq b$ and $0 \geq c \geq d$, $0 \leq -c \leq -d$ $\Rightarrow$ $-a c \leq -b d$ $\Rightarrow$ $a c \geq b d$.
Then, $u^{a - 1} (1 - u)^{b - 1} = u^{a - 1} (1 - u)^{a - 1} (1 - u)^{b - a} \leq (\frac{1}{4})^{a - 1} (1 - u)^{b - a} = (\frac{1}{4})^{a - 1}$ and $(a + b) u - a \geq -a$.
Thus we have $h(u) \geq -a (\frac{1}{4})^{a - 1}$.
For $f(a) = a (\frac{1}{4})^{a - 1}$, $f'(a) = (\frac{1}{4})^{a - 1} (1 - a \log(4)) < 0$, since $\log(4) > 1$ and $f(a)$ is a decreasing function of $a$, so the maximum $f(1) = 1 (\frac{1}{4})^{1 - 1} = 1 \Rightarrow h(u) \geq -1 \Rightarrow g(u) \geq 0$.
3.2. Suppose $a \geq b$.
Let $u = 1 - $t and then $t \in [\frac{b}{a+b}, 1]$, \begin{align} h(u) = h(1 - t) &= (1 - t)^{a - 1} t^{ b - 1} ((a + b) (1 - t) - a)\\ &= (1 - t)^{a - 1} t^{ b - 1} (a + b - (a + b) t - a)\\ &= (1 - t)^{a - 1} t^{ b - 1} (b - (a + b) t). \end{align} Method in 3.1 no longer works.
Thanks in advance.