A nonnegative, integrable, Lipschitz function $f$ satisfies $\lim \inf_{n \rightarrow \infty} \sqrt{n}f(n) = 0$

Let $f$ be a nonnegative integrable and Lipschitz function in $\mathbb{R}$ with Lipschitz constant $C$. Prove that $\lim \inf_{n \rightarrow \infty} \sqrt{n}f(n) = 0.$

The issue I have with this question is that why does the $\sqrt{n}$ matter? It seems to me like one approach would be to suppose that $f$ has finitely many zeroes. Let $N-1$ be the largest positive integer such that $f(N-1) = 0$, so we see that $$\inf\{\sqrt{N}f(N), \sqrt{N+1}f(N+1), ...\} = \alpha > 0.$$

Then we try to play around with the integrable and Lipschitz conditions to see if we can derive a contradiction. But even then I cannot settle on a good direction.

Suppose this is not true. Then there exists $r>0$ and $n_0$ such that $f(n) >\frac r {\sqrt n}$ for all $n \geq n_0$. Consider the intervals $(n,n+a_n)$ where $a_n=\frac 1 {\sqrt n \ln n}$ (for $n > 2$). For $x$ in this interval we have $f(x) \geq f(n)-|f(x)-f(n)|\geq f(n)- Ca_n$. So $\int f(x)dx \geq \sum_n \int_{(n,n+a_n)} f(x)dx \geq \sum_n [\frac {ra_n} {\sqrt n}-Ca_n^{2}] $. But $\sum_n \frac {ra_n} {\sqrt n}=\infty$ and $\sum_n a_n^{2}<\infty$. This proves that $\int f(x)dx=\infty$.