$f_{n}(x)= \frac{n^3 x^{3/2}}{ 1 + n^4 x^2}\}$ I think when $0 < x <1$ this expression is bounded by 1, am I correct? if so how can I prove it?
My trials:
At $x = 1,$ the following fraction $\{ \frac{n^3 x^{3/2}}{ 1 + n^4 x^2}\}$ becomes $\{\frac{n^3 }{ 1 + n^4}\}$ which is clearly less than 1 as $n^3 < 1 + n^4$ for all $n \in \mathbb{N}.$\
At $x = 0,$ the following fraction $\{ \frac{n^3 x^{3/2}}{ 1 + n^4 x^2}\}$ becomes $0$ which is less than 1.\
At $0 < x < 1,$ we have $x < 1,$ then $x^{3/2} < 1.$, so $n^3 x^{3/2} < n^3 < n^4 + 1.$ OTOH, $x^2 < 1$, then $1 + n^4 x^2 < 1 + n^4$ and so $\frac {1}{1 + n^4 x^2} > \frac{1}{ 1 + n^4}$, So it seems like in this case this term is not bounded by 1.
{A counterexample:}
if $n^{3}x^{3/2} \leq C (1+n^{4}x^{2})$ for all $n$ and for all $x$ then we can put $x =\frac 1 {n^{2}}$ to get $1 \leq 2C$ which is true. I am confused?
If you differentiate $f_n$ for $n \geq 1$ you have that the function attains its maximum at $x=\frac{\sqrt{3}}{n^2}$ on $[0,1]$