Let $n\in\mathbb{N}_{+}$ and $f(x), g(x)\in C^2[0,+\infty)$ be two functions satisfying $$f(0)=0,g(0)=1$$ $$f^{'}(x)=-\frac{1}{2}g(x)$$ $$g^{'}(x)=\frac{4n-x}{2x}f(x).$$ Prove that $$h(x)=\sqrt{\frac{x}{4n-x}}g^2(x)+\sqrt{\frac{4n-x}{x}}f^2(x)$$ has a lower bound of $\frac{1}{Cn}$ on $[\frac{1}{n-1},4n-4]$ where C is a constant.
Not very certain about the range where the bound holds. It's also okay if you proved a lower bound of $\frac{1}{Cn}$ on something like $[\frac{2}{n-1},4n-16\sqrt{n}]$.
I have already proved that $f(x)=e^{-\frac{1}{2}x}(L_{n}(x)-L_{n-1}(x))$ and $g(x)=e^{-\frac{1}{2}x}(L_{n}(x)+L_{n-1}(x))$ where $\\{L_n(x)\\}_{n=0}^{\infty}$ are the Laguerre polynomials, but I still have no idea.