Proving if $a_{k}\ge a_{k-1}+1$ then $1+\frac{1}{a_{0}}(1+\frac{1}{a_{1}-a_{0}})...(1+\frac{1}{a_{n}-a_{0}})\le \prod_{k=0}^{n}(1+\frac{1}{a_{k}})$ I've worked on this problem with the sequence $a_{k}$ being the natural numbers, that is $a_{k}=a_{k-1}+1$ and $a_{0}=1$. Over the naturals, $\prod_{k=0}^{n}(1+\frac{1}{a_{k}})$ can be proven to be $n+1$.
I've been able to recognize the obvious pattern in $\prod_{k=2}^{n}(1+\frac{1}{k-1})$ to be equal to $n$, but I do not know how to prove this rigorously.
That said, over the naturals, the inequality reduces to $n+1 \le n+1$. That got me thinking if the sequence in question could be expanded with the condition $a_{k}\ge a_{k-1}+1$ and specifically if this would not lead to the situation corresponding with RHS being greater than LHS.
Thank you for your help.


Given reals $a_0 > 0$ and $ a_k \geq a_{k-1} + 1$, show that $$ 1 + \frac{1}{a_0}\prod_{i=1}^{n}  ( 1 + \frac{1}{ a_i - a_0 })  \leq   \prod_{i=0}^n ( 1 + \frac{1}{a_i}).$$


Claim 1: $ \prod_{i=0}^n (1 + \frac{1}{a_i}) \leq \frac{a_{n+1}}{a_0} $
Proof: This follows from the telescoping inequalities
$$ \prod_{i=0}^n (1 + \frac{1}{a_i}) =  \prod_{i=0}^n \frac{a_i + 1}{a_i} \leq \prod_{i=0}^n \frac{ a_{i + 1} } { a_i} =  \frac{a_{n+1}}{a_0}. $$
Equality holds iff $a_i = i+a_0$.
Claim 2: $ 1 + \frac{1}{a_0} \prod_{i=1}^{n}  ( 1 + \frac{1}{ a_i - a_0 })  \leq \prod_{i=0}^n ( 1 + \frac{1}{a_i}).$
Proof by induction.
Base Case: $ 1 + \frac{1}{a_0} + \frac{1}{ a_0 (a_1 - a_0) } 
\leq 1 + \frac{1}{a_0} + \frac{1}{a_1} + \frac{1}{ a_0 a_1 } \\
\Leftrightarrow  a_1 \leq (a_0+1) (a_1 - a_0) \\
\Leftrightarrow a_0 ( a_1 - a_0 - 1) \geq 0 ,  $
which is true.
Induction step: Applying the induction hypothesis, we have
$$ 1 + \frac{1}{a_0} \prod_{i=1}^{k+1}  ( 1 + \frac{1}{ a_i - a_0 }) \leq 1 + ( 1 + \frac{1}{a_{k+1} - a_0 } ) \times \left[  \prod_{i=0}^k (1 + \frac{1}{a_i} ) - 1 \right] \\
  = ( 1 + \frac{1}{a_{k+1} - a_0 } ) \times \left[  \prod_{i=0}^k (1 + \frac{1}{a_i} ) \right] - \frac{1}{a_{k+1} - a_0} $$
We WTS that the RHS is $\leq ( 1 + \frac{1}{ a_{k+1} }) \prod_{i=0}^{k} ( 1 + \frac{1}{a_i}).$
Shifting terms around and factoring, this is equivalent to showing that
$$ \frac{a_0}{ (a_{k+1} - a_0 ) ( a_{k+1} ) } \times \left[ \prod_{i=0}^k ( 1 + \frac{1}{a_i})\right] \leq \frac{1}{ a_{k+1} - a_0 } .$$
This follows directly from the previous claim by replacing the term in the square brackets.
Again, equality holds iff $a_i = i+a_0$.

Note:

*

*The standard algebraic manipulations of the induction approach yielded the first claim directly. Assuming that this could be proved by induction in a direct manner, the first claim had to be true.

*The first claim could be strengthened to $ \prod_{i=0}^n (1 + \frac{1}{a_i}) \leq \frac{ a_{n} + 1 }{a_0}$.

