A problem about an integral inequality 
Let $f$ be a real continuous function on $[0,1]$ which satisfies the inequality 
  $$(f(t))^2 \le 1+2\int_{0}^{t}{f(s)\, ds} \quad\forall t \in [0,1].$$
  Prove that: 
  $$f(t) \le 1+t \quad\forall t \in [0,1].$$

 A: I revised my previous answer because the proof was not complete. Now I follow the approach suggested by SC30 (but avoiding the problematic at points where $g(s)=0$).    

We will show the following more general fact
  $$|f(t) |\le 1+t \quad\forall t \in [0,1].$$

For $\epsilon>0$, let 
$$g_{\epsilon}(x):=1+\epsilon+2\int_{0}^{x}f(s)\, ds\geq f^2(x)+\epsilon\geq \epsilon \quad\forall x \in [0,1].$$
then $|g_{\epsilon}'(x)|=2|f(x)|\leq2\sqrt{f^2(x)+\epsilon}\leq 2\sqrt{g_{\epsilon}(x)}$ and 
$$-1\leq \frac{g_{\epsilon}'(x)}{2\sqrt{g_{\epsilon}(x)}}\leq 1.$$
By integrating over $[0,t]$ with $t\in [0,1]$ we get
$$\int_{x=0}^t(-1)dx\leq \int_{x=0}^t\frac{g_{\epsilon}'(x)}{2\sqrt{g_{\epsilon}(x)}}dx\leq \int_{x=0}^t1dx.$$
that is
$$-t\leq\left[\sqrt{g_{\epsilon}(x)}\right]_0^t=\sqrt{g_{\epsilon}(t)}-\sqrt{1+\epsilon}\leq t.$$
Finally
$$|f(t)|\leq \sqrt{g_{\epsilon}(t)}\leq \sqrt{1+\epsilon}+|\sqrt{g_{\epsilon}(t)}-\sqrt{1+\epsilon}|\leq  \sqrt{1+\epsilon}+|t|.$$
Since $\epsilon$ is arbitrary, it follows that $|f(t)|\leq 1+|t|=1+t$.
A: Set $g(t)=1+2\int_{0}^t f(s)ds$, so $g(t)\geq f(t)^2\geq 0$. Now differentiating g we get $g'(t)=2f(t)\leq 2\sqrt{g(t)}$ for all $t\in [0,1]$ by the initial hypothesis ( we are allowed to do this since g is nonnegative). At this point we want to prove $f(t)\leq 1+t$ so it is the same as proving $f(t)-1\leq t$, but notice that $f(t)-1\leq \sqrt{g(t)}-1=\sqrt{g(t)}-\sqrt{g(0)}=\int_{0}^t (\sqrt{g(s)})'ds=\int \frac{g'(s)}{2\sqrt{g(s)}}ds\leq \int_{0}^t 1\cdot ds=t$, hence we have the desired inequality.
A: Set $g(t)=\int_0^t |\,f(s)|\,ds$. Then we have that $g\in C^1[0,1]$, $g(t)\ge 0$ and
$$
\big(g'(t)\big)^2=\big(f(t)\big)^2\le 1+2\int_0^tf(s)\,ds\le 1+2g(t), \quad \text{for all $t\in[0,1]$}.
$$
Hence
$$
\big(\sqrt{1+2g(t)}\big)'=\frac{g'(t)}{\sqrt{1+2g(t)}}\le 1, \quad \text{for all $t\in[0,1]$},
$$
and consequently, integrating the above over $[0,t]$, we obtain
$$
\sqrt{1+2g(t)}-\sqrt{1+2g(0)}=\sqrt{1+2g(t)}-1\le t,
$$
and thus
$$
\big(f(t)\big)^2\le 1+2\int_0^t f(s)\,ds=1+2g(t)\le (1+t)^2
$$
and finally
$$
f(t)\le |\,f(t)|\le |1+t|=1+t.
$$
A: This is a small variation of Yiorgos S. Smyrlis solution. We have
$$
 f^2(t) \le  1 + 2 \int_0^t  f(s) \, ds \le  1 + 2 \int_0^t \vert f(s) \vert \, ds 
$$
and therefore
$$ \tag{*}
 \vert f(t) \vert \le \left( 1 + 2 \int_0^t \vert f(s) \vert \, ds \right)^{1/2} \, .
$$
Denote the right-hand side of $(*)$ with $g(t)$. Then $g$ is
differentiable on $[0, 1]$ with $g(0) = 1$ and 
$$
 g'(t) = \frac 12 \left( 1 + 2 \int_0^t \vert f(s) \vert \, ds \right)^{-1/2} 2  \vert f(t) \vert
$$
Using $(*)$ we conclude that $g'(t) \le 1$ and thus
$$
 f(t) \le \vert f(t) \vert \le g(t) \le 1 + t \, .
$$
