$\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$ 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 $x > 100$

Show this product is equal to:
$$\frac{4^{5000}}{\binom{10000}{5000}}$$
Then use the inequality about the central binomial coefficient:
$$\frac{4^n}{\sqrt{4n}}\leq\binom{2n}{n}\leq\frac{4^n}{\sqrt{3n+1}}$$
Setting $n=5000$ this gives:
$$\frac{4^{5000}}{\sqrt{20000}}<\binom{10000}{5000}<\frac{4^{5000}}{\sqrt{15001}}$$
Or
$$122<\sqrt{15001}<\frac{4^{5000}}{\binom{10000}{5000}}<\sqrt{20000}<142$$
