Supremum question with Arithmetic-Geometric means Let $n\ge2$ fixed. How do you find the least $\alpha(n)$ such that, for any numbers $0<a_1,...,a_n$ and $0\le b_1,...,b_n\le1/2$, with $\sum_{i=1}^n a_i=\sum_{i=1}^n b_i=1$, we have $\prod_{i=1}^na_i\le\alpha(n)\sum_{i=1}^na_ib_i$.
I was thinking use the generalized AM-GM inequality, but i can't finish it.
 A: Let $n> 2$.  WLOG let $a_i$ be ordered in non-decreasing manner.  Then, $\sum_{i=1}^n a_i b_i \geqslant \frac12(a_1+a_2)$ so it is enough to consider the inequality
$$\prod_{i=1}^n a_i \leqslant \alpha(n) \cdot \frac{a_1+a_2}2$$
Once $a_1, a_2$ are fixed, we have $\prod_{i> 2} a_i$ maximised (by AM-GM) when all higher $a_i$ are identical, i.e. we may set $\displaystyle a_{i>2} = \frac{1-a_1-a_2}{n-2}$ for the optimum.  
For convenience, let $p^2 = a_1 a_2, 2s = a_1+a_2$.  Then it is enough to consider the inequality
$$p^2\cdot\left(\frac{1-2s}{n-2} \right)^{n-2} \leqslant \alpha(n)s$$ 
where we need to optimise for $p, s$, with $0 \leqslant p \leqslant s \leqslant \frac12$.  Clearly reducing $s$ makes the LHS larger and the RHS smaller. hence making the inequality tighter.  Similarly increasing $p$ makes the LHS larger.  Hence we should set $s=p$, i.e. $a_1 = a_2=s=p$ in the optimum.  With this, we have
$$s\left(\frac{1-2s}{n-2} \right)^{n-2} \leqslant \alpha(n)$$
Using univariate calculus (or AM-GM), it is easily seen LHS has a maximum when $s=\frac1{2n-2}$.  Interestingly this sets $a_1 = a_2$ at half the value as all the rest. Using this, we get $\alpha(n) \geqslant \dfrac1{2(n-1)^{n-1}}$.

P.S. $n=2$ is left for you to work out...
