Another inequality related to (weighted) AM-GM ...

Question

Let $w_i \in [0,1], \sum_i w_i = 1$ (i.e. weights); $\beta_i \in (0,1)$; and $N \ge 1$.

We can show that: $$ [A]\qquad1 - \prod_i {(1 - \beta_i)}^{N\,w_i} \quad\ge\quad 1 - \sum_i w_i {(1 - \beta_i)}^N\qquad[B] $$ by the weighted AM-GM inequality, using a substitution: $x_i = (1 - \beta_i)^N$.

However, I'd like to know if we can show an inequality w.r.t. [A], [B], and new [C]: $$ 1 - {(1 - \sum_i w_i\,\beta_i)}^N \qquad[C] $$ From some basic simulation, I think that $[A] \ge [C] \ge [B]$, but I'm not sure if this always holds or how to show this analytically. TIA.

Answer 1

$[A]\geq[C]$: Using Jensen's inequality

\begin{align} 1 - \prod_i(1 - \beta_i)^{Nw_i}&\geq 1 - (1 - \sum_i w_i\beta_i)^N &\iff\\ \prod_i(1 - \beta_i)^{Nw_i} &\leq (1 - \sum_i w_i\beta_i)^N &\iff\\ N\sum_i w_i\log(1-\beta_i)&\leq N\log(1 - \sum_iw_i\beta_i), \end{align}

which follows from $f(t)=N\log(1-t)$ being concave on $(0,1)$ since

$f''(t)=-\frac{1}{(1-t)^2}\leq 0$ for $0<t<1$.

$[C]\geq[B]$:

\begin{align} 1 - (1 - \sum_i w_i\beta_i)^N&\geq 1 - \sum_i w_i(1-\beta_i)^N &\iff\\ (1 - \sum_i w_i\beta_i)^N&\leq \sum_i w_i(1-\beta_i)^N,\end{align}

which follows from $f(t)=(1-t)^N$ being convex on $(0,1)$ since

$f''(t) = N(N-1)(1-t)^{N-2}\geq 0$ for $0 < t < 1$.