Inequality for a function 4 Let $u:[0,+\infty)\to\mathbb R^+$ be a bounded positive function such that
$$u(t)\leq \int_0^t\left(-\frac{1}{\sqrt N}u(s)+\frac{1}{N}\right)ds +\frac{1}{N^{\frac{1}{4}}}$$  for every $t\geq 0$, where $N\in\mathbb N$.
Is it correct that 
$$u(t)\leq\frac{1}{\sqrt N}+\frac{1}{N^{\frac{1}{4}}}$$
for every $t\geq 0$?
How could I prove that?

The inequality does not hold. As a counterexample, consider
$$ u(t) = \begin{cases}1/2,  &0\leq t < 8 \\ 3, &8\leq t<9 \\ 1/2, &9\leq t<\infty\end{cases} $$
Then  one can show that $u$ satisfies the first inequality for $N = 1$:
$$ \int_0^t\left(-\frac{1}{\sqrt N}u(s)+\frac{1}{N}\right)ds +\frac{1}{N^{\frac{1}{4}}} = 1 + \int_0^t(1-u(s))\,ds $$
and 
$$1 + \int_0^t(1-u(s))\,ds = \begin{cases}1 + \frac 1 2t,  &0\leq t < 8 \\ 5 - 2(t-8), &8\leq t<9 \\ 3 + \frac{1}{2}(t-9), &9\leq t<\infty\end{cases}, $$
so $u(t) \leq 1 + \int_0^t(1-u(s))\, ds$. But $$u(8) = 3 > 2 = \frac 1 {\sqrt N} + \frac 1 {N^{1/4}}.$$ 
