Inequality of functions I would like to show that $h_n(x) = \frac{n\sqrt{n}x}{1+n^2x^2} \leq \frac{1}{\sqrt{x}} = f(x)$ for every natural number $n$ and for every $x \in (0,1)$. It's obvious when $n = 0$ as $h_0(x) = 0$ for every $x \in (0,1)$. I am, however, struggling to show that the inequality holds for other $n$ values. I tried to play with $f(x)$ by changing the inequality to $\frac{n\sqrt{n}x}{1+n^2x^2} \leq \frac{n^2\sqrt{x}x}{n^2x^2}$, but I am getting nowhere. Any help will be greatly apprecaited. 
 A: Generalise the inequality and let $n,x\in\mathbb{R}^+$. Clear denominators to get
$$n\sqrt{n}x\sqrt{x}\leq 1+n^2 x^2.$$ 
If we now let $t=nx$, we have to prove that for all reals $t>0$ the inequality $t\sqrt{t}\leq 1+t^2$ holds. This is equivalent to 
$$t^3\leq 1+2t^2+t^4\Leftrightarrow t^3(t-1)+2t^2+1\geq 0.$$
Obviously, for $t\geq 1$, the left hand side is positive. If $0<t<1$, however, we have $0>t^3(t-1)>-1$, so that the left hand side is greater or equal to $2t^2\geq 0$. Thus, the inequality is proved for all $n,x\in\mathbb{R}^+$, which includes your inequality as a special case.
A: Young's inequality with $p=4$ and $q=4/3$ gives
$$
 1 \cdot (nx)^{3/2} \le \frac 14 + \frac 34 (nx)^2 < \frac 34 (1 + (nx)^2)
$$
which gives the slightly better estimate
$$
\frac{n\sqrt{n}x}{1+n^2x^2} < \frac{3}{4\sqrt{x}}
$$
for arbitrary $n, x > 0$.
A: Clearing out the denominators, what you are trying to show is
$$(nx)^{3/2} \leq 1 + (nx)^2$$
If $nx > 1$, then raising $nx$ to a higher power will only enlarge it, so since ${3 \over 2} < 2$ one has
$$(nx)^{3/2} < (nx)^2 < 1 + (nx)^2$$
On the other hand, if $nx \leq 1$, then any power of 
$nx$ is also at most $1$, so we have
$$(nx)^{3/2} \leq 1 < 1 + (nx)^2$$
Either way, the desired equation holds, so we're done.
