Does this inequality hold? And can it be shown? If $x>1$ and $y$ is some positive integer greater than 1, then is it always true that,
$\frac{x}{x+1} \prod_{i=1}^y \left( 1 + \frac{1}{xy+2i-2} \right) < 1$?
Graphing the expression seems to suggest that it is. But can this be shown?
 A: I believe your inequality holds.
Let's define
$$
f_n(x) = \frac{x}{x+1} \prod_{k=1}^n
\biggl( 1 + \frac{1}{nx + 2k - 2} \biggr),
$$
and phrase your question as asking whether $f_n(x) < 1$ for all $x > 1$.
Now, I don't like the term in the product, so I'm going to try to estimate it by something nicer. We all know that $k-1 \leq 2k-2$ holds when $k \geq 1$, so $nx + 2k - 2 \geq nx + k -1$ for $k \geq 1$. Note also that
$$
1 + \frac{1}{nx +k - 1} = \frac{nx + k}{nx + k - 1}.
$$
Then we get the upper bound
$$
f_n(x) \leq 
\frac{x}{x+1} \prod_{k=1}^n
\biggl( 1 + \frac{1}{nx + k - 1} \biggr)
=
\frac{x}{x+1} \frac{\prod_{k=1}^n nx + k}{\prod_{k=1}^n nx + k - 1}.
$$
Now, the product there telescopes - the $k$-th term on the bottom cancels out against the $(k+1)$-th term on the top - so we're left with the estimate
$$
f_n(x) \leq \frac{x}{x+1} \frac{nx + n}{nx} = 1.
$$
Looking back at the statement of the problem, we notice that $n \geq 2$, so there's always a $k$ for which we get a strict inequality in our first estimate, so we have proved the strict inequality $f_n(x) < 1$ for all $x > 1$ (actually $x > 0$, by staring at this for a bit).
