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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!

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  • $\begingroup$ Stope please to change the problem and open another topic with your new problem. $\endgroup$ Oct 17, 2020 at 8:16
  • $\begingroup$ Also, you can post your new problem without deleting of the first. I think it would be better. $\endgroup$ Oct 17, 2020 at 8:34

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

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  • $\begingroup$ Apologize for typing the conclusion wrongly. Would you do it again? $\endgroup$
    – user823011
    Oct 17, 2020 at 8:07
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    $\begingroup$ @user823011 But it's really another problem. It's not faire. Restore please the previous problem and open another topic with your new problem. $\endgroup$ Oct 17, 2020 at 8:09
  • $\begingroup$ @user823011 All problems in math are similar. Just post your new problem in another topic. By the way, I solved your new problem too. Do it. $\endgroup$ Oct 17, 2020 at 8:27
  • $\begingroup$ However the system forbids me froming posting the new one due to the similarity, and I cannot delete this one now. :( $\endgroup$
    – user823011
    Oct 17, 2020 at 8:28
  • $\begingroup$ @user823011 OK. Post your new problem without deleting of the first. I think it would be better. $\endgroup$ Oct 17, 2020 at 8:30

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