Help with a dominated convergence theorem problem The problem is to find an integrable function that bounds $f(x)={\dfrac{{{n^{\frac{3}{2}}}x}}{{1 + {n^2}{x^2}}}}$ on $[0,1]$ so we can calculate $\displaystyle\int_0^1 {\frac{{{n^{\frac{3}{2}}}x}}{{1 + {n^2}{x^2}}}dx} $ using dominated convergence theorem. I know by taking derivative with respect to $n$ we can find the maximum point $n=\frac{\sqrt 3}{x}$  which leads to $f(x)$ is bounded by $x^{-\frac{1}{2}}$, which is integrable on $[0,1]$, i.e. $|f(x)|\le x^{-\frac{1}{2}}$ on $[0,1]$.
My question is, is there nicer solution without taking derivatives? I have tried an hour using inequalities like $1+n^2x^2\ge 2nx$ or substituting $1$ in the denominator by $x$, or $x^\frac{1}{2}$, but all failed. I am so tired so I post this question here and hope someone can help.
I used derivative in my homework but my professor commented "try not using derivatives".
Thank you!

By AM-GM: $\dfrac{1+n^2x^2}{n^{3/2}x^{3/2}} = \dfrac{1}{(nx)^{3/2}} + (nx)^{1/2} = \dfrac{1}{(nx)^{3/2}} + \dfrac{(nx)^{1/2}}{3} + \dfrac{(nx)^{1/2}}{3} + \dfrac{(nx)^{1/2}}{3}$ 
$\ge 4 \sqrt[4]{\dfrac{1}{(nx)^{3/2}}\cdot \dfrac{(nx)^{1/2}}{3}\cdot\dfrac{(nx)^{1/2}}{3}\cdot\dfrac{(nx)^{1/2}}{3}}$ $= \dfrac{4}{3^{3/4}}$. 
Hence, $\dfrac{n^{3/2}x^{3/2}}{1+n^2x^2} \le \dfrac{3^{3/4}}{4}$, and thus, $\dfrac{n^{3/2}x}{1+n^2x^2} \le \dfrac{3^{3/4}}{4x^{1/2}}$, which is integrable on $[0,1]$.
