Q:

Proof of a Integral Inequality with Calculus

From the Fall 2016 UC Berkeley preliminary exam
Let $x:[a,b] \rightarrow \mathbb{R}$ and $f:[a,b] \rightarrow \mathbb{R}$ be non-negative continuous functions satisfying
$$x^2(t) \leq 1+ \int_a^t f(s)x(s)ds$$
for $a\leq t\leq b$. Show that 
$$x(t) \leq \frac{1}{2}\int_a^tf(s)ds$$
for $a\leq t\leq b$.
Solution: Let
$$y(t)=1+\int_a^t f(s)x(s)ds$$
so that
$$x(t) \leq \sqrt{y(t)}$$
Then
$$y'(x)=f(t)x(t)\leq f(t)\sqrt{y(t)}$$
that is,
$$\frac{y'}{\sqrt{y}}=(2y^{\frac{1}{2}})' \leq f$$
Integrating this gives (this is the step I'm confused about)
$$x(t) \leq \sqrt{y(t)} \leq 1+\frac{1}{2}\int_a^tf(s)ds$$
When you integrate the second to last expression from $a$ to $t$, specifically when you integrate $f$, where does that $1 +$ in front pop out from?

A:

From $\frac{y'(s)}{\sqrt{y(s)}} \leq f(s)$ we get
$ \int_a^t \frac{y'(s)}{\sqrt{y(s)}} \leq \int_a^t f(s) ds$. 
The integral on the left is $=2\sqrt{y(s)}|_a^t=2\sqrt{y(t)}-2\sqrt{y(a)}=2\sqrt{y(t)}-2$, since $y(a)=1$.

