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$\textbf{Question:}$ Prove using induction that the following inequality holds for all positive integers $n$:

$$\dfrac{(1+a_1)(1+a_2)\cdots(1+a_n)}{1+a_1a_2\cdots a_n}\leq 2^{n-1},$$ where $a_1,a_2,\dots ,a_n\geq 1$.

$\textbf{Attempted (Incorrect) Solution:}$ Base Case: $\dfrac{(1+a_1)}{(1+a_1)}=1=2^0=2^{1-1}$. Assume we have $\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)}{1+a_1a_2\cdots a_k}\leq 2^{k-1}$ for some positive integer $k$. Clearly, $\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)}{1+a_1a_2\cdots a_ka_{k+1}}\leq\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)}{1+a_1a_2\cdots a_k}\leq 2^{k-1}$, so $\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)(1+a_{k+1})}{1+a_1a_2\cdots a_ka_{k+1}}=\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)}{1+a_1a_2\cdots a_ka_{k+1}}+\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)a_{k+1}}{1+a_1a_2\cdots a_ka_{k+1}}\leq 2^{k-1}+\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)a_{k+1}}{1+a_1a_2\cdots a_ka_{k+1}}\leq 2^{k-1}+ \dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)a_{k+1}}{a_1a_2\cdots a_ka_{k+1}}\leq 2^{k-1}+ \dfrac{2^k(a_1a_2\cdots a_k)a_{k+1}}{a_1a_2\cdots a_ka_{k+1}}=2^{k-1}+2^k=3\cdot2^{k-1}\not\leq 2^k.$

I've tried a couple of different approaches, but none of them have panned out. I feel like I might be missing some key insight/trick. Any hints/suggestions would be greatly appreciated. Thanks!

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Hint:

With $b_k = a_k a_{k+1} \ge 1, \quad c_k = a_k + a_{k+1}-1 \ge 1$,

$(1+a_k)(1+a_{k+1}) = 1+(c_k+1) + b_k \le (1 + b_k) + (1 + b_k c_k)$
and $b_kc_k \ge b_k = a_k a_{k+1}$

So split the LHS of the inductive step into two sums...

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