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.

Proof: We will use the following bound:
$$\frac{1}{x^2 + \ln x} < \frac{1}{x^2} - \frac{3}{2x^4}, \quad \forall x\ge 5.$$
proof: It suffices to prove that $f(x) = x^2 + \ln x - \frac{2x^4}{2x^2-3} > 0$ for $x\ge 5$.
We have $f'(x) = \frac{4x^4+6x^2+9}{x(2x^2-3)^2} > 0$ for $x\ge 5$. Note also that
$f(5) > 0$. The desired result follows.
With the bound above, noting also that $\ln (1+x) \le x$ for $x\ge 0$, we have
\begin{align}
&\prod_{n=1}^\infty \Big(1 + \frac{1}{n^2 + \ln n}\Big)\\
=\ & \prod_{n=1}^4 \Big(1 + \frac{1}{n^2 + \ln n}\Big)
\cdot \mathrm{exp}\left(\sum_{n=5}^\infty \ln\Big(1 + \frac{1}{n^2 + \ln n}\Big) \right)\\
\le\ & \prod_{n=1}^4 \Big(1 + \frac{1}{n^2 + \ln n}\Big)
\cdot \mathrm{exp}\left(\sum_{n=5}^\infty \Big(\frac{1}{n^2} - \frac{3}{2n^4}\Big) \right)\\
=\ & \prod_{n=1}^4 \Big(1 + \frac{1}{n^2 + \ln n}\Big)
\cdot \mathrm{exp}\left(\frac{2689}{13824}+\frac{\pi^2}{6}-\frac{\pi^4}{60} \right)\\
< \frac{7}{2}
\end{align}
where we have used $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$
and $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$.
