First note:
$$\int_0^x f(x) \, \textrm{d}x = \int 1_{[0,x]}(t) f(t) \, \textrm{d}t.$$
Then:
$$\left (\int 1_{[0,x]}(t) f(t) \, \textrm{d}t \right )^2 \leq \int 1_{[0,x]}(t)^2 \, \textrm{d}t \cdot \int_0^x f(t)^2 \, \textrm{d}t.$$
By Cauchy-Bunyakovski-Schwarz.
Thanks to the hint of @Saad in comments, I am now able to finish the proof.
First I separate the summand in the integral into two parts (the first one will be relevant when $t$ is small, the second one when $t$ is large):
$$\exp\left(-\frac{t^2}{2(1+t)}\right)\le \exp\left(-\frac{t^2}{4\max(1,t)}\right)\le e^{-t^2/4}+e^{-t/4}$$
Now I can use the hint in the following way:
$$\mathbb{E} Z=\int_0^\infty\mathbb{P}[Z\ge t]\le \int_0^\infty \min\left(C \exp\left(-\frac{t^2}{2(1+t)}\right),1\right)\,dt\le A+B$$
with:
$$A=\int_0^\infty \min(C e^{-t^2/4},1)\,dt$$
$$B=\int_0^\infty \min(C e^{-t/4},1)\,dt$$
For $B$ the crossing point of the two regimes is for $t=4 \log C$, hence:
$$B\le 4\log C+\int_{4\log C}^\infty C e^{-t/4}\,dt=4(\log C+1)$$
For $A$ it happens at $t=2\sqrt{\log C}$, so that:
$$A\le 2\sqrt{\log C}+\int_{2\sqrt{\log C}}^\infty Ce^{-t^2/4}\,dt$$
And the integral can be upper bounded as follows:
$$\int_{2\sqrt{\log C}}^\infty Ce^{-t^2/4}\,dt=\sqrt{2} C\int_{\sqrt{2\log C}}^\infty e^{-t^2/2}\,dt\le 2\sqrt{\pi}$$
where I used the standard inequality:
$$\int_x^\infty e^{-t^2/2}\, dt\le \sqrt{2\pi} e^{-x^2/2}$$
Best Answer
Showing that $$\int_{t\sqrt{n}}^{\pi\sqrt{n}/2} e^{-\phi^2/2}\mathrm{d}\phi \leq \sqrt{2\pi}e^{-t^2(n+2)/2}$$ means (after integration and simplifying) that we need to show that $$F(n,t)=2 e^{-\frac{1}{2} (n+2) t^2}+\text{erf}\left(\frac{\sqrt{n} t}{\sqrt{2}}\right)-\text{erf}\left(\frac{\pi \sqrt{n}}{2 \sqrt{2}}\right) \geq 0$$ Checking at the bounds, it is true since $$F(n,0)=2-\text{erf}\left(\frac{\pi \sqrt{n}}{2 \sqrt{2}}\right)>0 \qquad \text{and} \qquad F\left(n,\frac{\pi }{2}\right)=2 e^{-\frac{\pi ^2}{8} (n+2)}>0$$ The first derivative $$F'(n,t)=e^{-\frac{n+2}{2} t^2} \left(\sqrt{\frac{2n}{\pi }} e^{t^2}-2 (n+2) t\right)$$ cancels only once at $$t_*=\sqrt{-\frac{1}{2} W\left(-\frac{n}{\pi (n+2)^2}\right)}$$ and the second derivative test shows that this is a maximum.
So, it is true.