$\frac{b_1b_2\cdots b_n}{a_1a_2\cdots a_n}\leq\frac{B}{A}$, prove $\frac{(b_1+1)(b_2+1)\cdots (b_n+1)}{(a_1+1)(a_2+1)\cdots (a_n+1)}\leq\frac{B}{A}$

Question

Let $n\in\mathbb{N}^+$, $a_i,b_i, A, B\in\mathbb{R}^+$, $a_i\leq b_i, a_i\leq A\ (i=1,2,\cdots, n),$ and $$\dfrac{b_1b_2\cdots b_n}{a_1a_2\cdots a_n}\leq\dfrac{B}{A}. $$ Prove that$$\frac{(b_1+1)(b_2+1)\cdots (b_n+1)}{(a_1+1)(a_2+1)\cdots (a_n+1)}\leq\frac{B}{A}$$

I tried to use induction and have proved $n=1$ case, but it seems it has some difficulty to go from $n$ to $n+1$ (my method is to let $a_1a_2, b_1b_2$ be a whole, but to do this we have to assume $a_1a_2\leq A$).

Apprecaite a valid and complete proof!

Answer 1

Since $b_i\geq a_i$, we obtain $$\frac{b_i}{a_i}\geq\frac{1+b_i}{1+a_i}$$ and $$\frac{B}{A}-\frac{\prod\limits_{i=1}^n(1+b_i)}{\prod\limits_{i=1}^n(1+a_i)}\geq\frac{\prod\limits_{i=1}^n b_i}{\prod\limits_{i=1}^n a_i}-\frac{\prod\limits_{i=1}^n(1+b_i)}{\prod\limits_{i=1}^n(1+a_i)}\geq0.$$