<s>[INST] Proving the inequality $\prod_{n=1}^\infty \left( 1+\frac1{n^2+\ln n} \right) < \frac72$ I am struggling to prove
$$\prod_{n=1}^k \left( 1+\frac1{n^2+\ln n} \right) < \frac72$$
For all $k \geq 1$. Clearly, this is true for $k=1$, so it suffices to show

$$\prod_{n=1}^\infty \left( 1+\frac1{n^2+\ln n} \right) < \frac72$$

The obvious way to show this would be to evaluate the product directly, though I don't think that is feasible. I am not sure how else to approach this product. This would be a (much) tighter upper bound than 
$$\exp\left ({\sum_{n=1}^k} \frac1{n^2+\ln n} \right )$$
Which comes from the monotone convergence theorem.
 [/INST] I have not full solution to this problem.
I want to prove by induction that:
$$ \prod_{n=1}^{k} \left(1 + \frac{1}{n^2 + \ln{n}}\right) < \frac{7}{2} \frac{(k+1)^2 + \ln(k+1)}{(k+2)^2 + \ln(k+2)} < \frac{7}{2}$$
It's obvious for $k=1$.
Suppose we proved for $k$:
$$ \prod_{n=1}^{k + 1} \left(1 + \frac{1}{n^2 + \ln{n}}\right) < \frac{7}{2} \left(1 + \frac{1}{(k+1)^2 + \ln{(k+1)}}\right)  \frac{(k+1)^2 + \ln(k+1)}{(k+2)^2 + \ln(k+2)}$$
So we need to prove that:
$$ \left(1 + \frac{1}{(k+1)^2 + \ln{(k+1)}}\right)  \frac{(k+1)^2 + \ln(k+1)}{(k+2)^2 + \ln(k+2)} < \frac{(k+2)^2 + \ln(k+2)}{(k+3)^2 + \ln(k+3)} \\
\frac{(k+1)^2 + \ln(k+1) + 1}{(k+2)^2 + \ln(k+2)} < \frac{(k+2)^2 + \ln(k+2)}{(k+3)^2 + \ln(k+3)}
$$ 
But I don't know how to prove the last inequality, which according to Wolfram is true for $k > -1$.
Probably I will improve my answer after a while.
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