$\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$ Prove that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$
saw some similar problems like
show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true
but didn't manage to get 115. I could get a weaker conclusion of $x>100$ though.
\begin{align}
x^2 &= \left(\frac 21 \times \frac 21\right) \times \left(\frac 43 \times \frac 43\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{9999}\right) \\
&\ge \left(\frac 21 \times \frac 32\right) \times \left(\frac 43 \times \frac 54\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10001}{10000}\right) \\
&= 10001
\end{align}
so $x > 100$

You can make the problem more general since you have
$$S_p=\frac{\prod_{n=1}^p (2n) }{\prod_{n=1}^p (2n-1) }=\frac{2^p \Gamma (p+1) } {\frac{2^p \Gamma \left(p+\frac{1}{2}\right)}{\sqrt{\pi }} }=\sqrt{\pi }\,\,\frac{ \Gamma (p+1)}{\Gamma \left(p+\frac{1}{2}\right)}$$ Take logarithms, use Stirling approximation and continue with Taylor series for large $p$
$$A=\log (\Gamma (p+1))-\log \left(\Gamma \left(p+\frac{1}{2}\right)\right)$$
$$A=\frac{1}{2}\log (p)+\frac{1}{8 p}-\frac{1}{192
   p^3}+O\left(\frac{1}{p^5}\right)$$
$$\frac{ \Gamma (p+1)}{\Gamma \left(p+\frac{1}{2}\right)}=e^A=\sqrt{p}\left(1+\frac{1}{8 p}+\frac{1}{128 p^2}+O\left(\frac{1}{p^3}\right) \right)$$
$$S_p >\sqrt{p\pi}\left(1+\frac{1}{8 p}\right) $$ Using it for $p=5000$ gives $125.334547017$ while the exact value is $125.334547056$
