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

Looking at your method, you've actually proven $\frac{3}{4}x^2 > 10000$ as well – just, instead of writing $\frac{2}{1} \geqslant \frac{3}{2}$, you can incorporate that $\frac{3}{4}$ here, and you'll have exactly $\frac{3}{2} = \frac{2}{1} \cdot \frac{3}{4}$. Now we can deduce that

$$ x > \sqrt{\frac{40000}{3}} \approx 115.47 > 115$$

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

$$122<\sqrt{15001}<\frac{4^{5000}}{\binom{10000}{5000}}<\sqrt{20000}<142$$

We have $$\dfrac{2}{1}\cdot \dfrac{4}{3}\cdots \dfrac{10000}{9999}= \dfrac{10000!!}{9999!!}= \dfrac{2^{5000}\cdot 5000!}{\dfrac{10000!}{2^{5000}\cdot 5000!}}= 2^{10000}\cdot \dfrac{\left ( 5000! \right )^{2}}{10000!}$$ Using Sterling's approximation $n!= \sqrt{2n\pi}\left ( \dfrac{n}{e} \right )^{n}$ then we have $$x\cong 2^{10000}\cdot \dfrac{\left ( \sqrt{10000\pi}\left ( \dfrac{5000}{e} \right )^{5000} \right )^{2}}{\sqrt{20000\pi}\left ( \dfrac{10000}{e} \right )^{10000}}= 2^{10000}\cdot \dfrac{50\sqrt{2\pi}}{2^{10000}}= 50\sqrt{2\pi}$$ Another solution is $$x= \frac{2^{10000}}{\dbinom{10000}{5000}}= \frac{2^{10000}}{\dfrac{2^{10001}}{\pi}\int_{0}^{\infty}\dfrac{{\rm d}x}{\left ( x^{2}+ 1 \right )^{5001}}}= \frac{\pi}{2\int_{0}^{\infty}\dfrac{{\rm d}x}{\left ( x^{2}+ 1 \right )^{5001}}}$$ Then by the transformation $\int_{0}^{\infty}f\left ( x \right ){\rm d}x= \int_{0}^{1}\dfrac{f\left ( \dfrac{x}{1- x} \right )}{\left ( 1- x \right )^{2}}{\rm d}x$ we have $$\int_{0}^{\infty}\frac{{\rm d}x}{\left ( x^{2}+ 1 \right )^{5001}}= \int_{0}^{1}\frac{{\rm d}x}{\left ( \left ( \dfrac{x}{1- x} \right )^{2}+ 1 \right )^{5001}\left ( 1- x \right )^{2}}= \int_{0}^{1}\frac{\left ( x- 1 \right )^{10000}}{\left ( 2x^{2}- 2x+ 1 \right )^{5001}}{\rm d}x=$$ $$= \int_{0}^{1}\frac{\left ( \dfrac{x^{2}- 2x+ 1}{2x^{2}- 2x+ 1} \right )^{5000}}{2x^{2}- 2x+ 1}{\rm d}x$$ We can doublecheck the result then in Wolfram|Alpha or Mathematica.