Well, it's not a hard problem . It's obvious that the product of S multiplied by R is 1/225. Also, every single item in R is less than the corresponding item in S. (e.g. 2/3 > 1/2 ; 4/5 >3/4 ) So R is less than S. Then we've done.

See, $$R\cdot S=\frac{1}{225}=\left(\frac{1}{15}\right)^2$$Then we will prove $A<B$. If $x$ positive integer, $$(x+1)(x-1)=x^2-1<x^2\Longleftrightarrow \frac{x-1}{x}<\frac{x}{x+1}$$such that, $$\prod_{x=2}^{224} \frac{x-1}{x}<\prod_{x=2}^{224} \frac{x}{x+1}$$We get $A<B\Longleftrightarrow A\ne B$. Thus from $RS=\dfrac{1}{15^2}\Longleftrightarrow A<\dfrac{1}{15}<B$. QED. Rafa171206

We just need to prove that $R<S$, since $RS=\frac1{225}$ and $R<S$ would imply that $R<\frac1{15}$, since $\frac1{15}$ is the geometric mean of $R$ and $S$. But this is true, since $\frac x{x+1}<\frac{x+1}{x+2}\Leftrightarrow x^2+2x<x^2+2x+1$ (obvious), and thus $R=\prod_{n=1}^{112}\frac{2n-1}{2n}<\prod_{n=1}^{112}\frac{2n}{2n+1}=S$. $\square$

An alternative is to note that $\frac{2x}{2x+1} > \sqrt{\frac{2x-1}{2x+1}}$ for all positive $x$, so $S > \sqrt{\frac{1}{3}\cdot\frac{3}{5}\cdots\frac{223}{225}} = \frac{1}{15}$. The inequality for $R$ now follows from $RS = \frac{1}{225}$.