Suppose $x:[0,\infty) \rightarrow [0,\infty)$ is continuous and $x(0)=0$ and if
$$ x(t)^2 \leq 2+ \int _0 ^tx(s)ds $$
for all $t\geq 0$, then which of the following is true?
1) $x(\sqrt 2)\in [0, 2]$
2) $x(\sqrt 2)\in [0, \frac{3}{\sqrt 2}]$
3) $x(\sqrt 2)\in [\frac{5}{\sqrt 2}, \frac{7}{\sqrt 2}]$
4) $x(\sqrt 2)\in [10,\infty]$
I do using equality of the above relation and answer I got $1/ \sqrt 2$
Let $$y(t):=2+\int_0^t{x(s)}ds\geq 2$$ Then from the intgral inequality we have $$\dot{y}=x(t)\leq \sqrt{y}$$ or equivalently $$\frac{d}{dt}\left(2{y^{1/2}(t)}-t\right)\leq 0$$ which yields $$2y^{1/2}(t)-t\leq 2y^{1/2}(0)=2\sqrt{2}$$ i.e. $$y^{1/2}(t)\leq \frac{t}{2}+\sqrt{2}$$ From the initial inequality we have $$x(t)\leq \left(2+\int_0^t{x(s)ds}\right)^{1/2}=y^{1/2}(t)\leq \frac{t}{2}+\sqrt{2} $$ For $t=\sqrt{2}$ we obtain from above $$x(\sqrt{2})\leq \frac{3}{\sqrt{2}}$$