I have faced with the following problem.
Assume $(x_1,x_2,\dots,x_m)$ are arbitrary positive real numbers and we have the ordered sequence $d_1\le d_2\le \cdots \le d_m$ of positive real numbers. define
$$A = (1+\frac{b_1}{a_1}x_1)^{d_1a_1}(1+\frac{b_1}{a_2}x_2)^{d_2a_2}\cdots(1+\frac{b_1}{a_m}x_m)^{d_ma_m}$$
and
$$B = (1+\frac{b}{a}\sum_{i=1}^{m}x_i)^{d_1a}(1+\frac{b}{a}\sum_{i=2}^{m}x_i)^{(d_2-d_1)a}\cdots (1+\frac{b}{a}\sum_{i=k}^{m}x_i)^{(d_k-d_{k-1})a}\cdots \\ (1+\frac{b}{a}x_m)^{(d_m-d_{m-1})a}$$
where $\forall i: 0 \le a_i \le C$ , $0 \le b_1 \le C$, $0 \le b \le C$ and $0 \le a \le C$ for a known value of parameter $C$ and there are constraints $b_1+\sum_{i=1}^{m}a_i \le C$ and $a+b \le C$. I am trying to find the conditions on $a_i$ which makes $B \ge A$.
Can I use the Karamata's inequality? If I can show the sequence $\left[(1+\sum_{i=1}^{m}x_i)^{d_1a}, (1+\sum_{i=2}^{m}x_i)^{(d_2-d_1)a},\cdots,(1+x_m)^{(d_m-d_{m-1})a}\right]$ majorizes $\left[(1+x_1)^{d_1a_1},(1+x_2)^{d_2a_2},\cdots,(1+x_m)^{d_ma_m}\right]$ then I can use convex function $f(x)=\ln(x)$ to arrive at desired result. Is it the correct course of action? Is there any other way, any other inequality?
Any suggestion would be appreciated.
