Differential equation monotonicity:$(1+f(x))f''(x)=1+x$ 
$f(x)$ is defined $\forall x \geq 0$ and has a continuous derivative. It satisfies $f(0)=1,f'(0)=0$ and $(1+f(x))f''(x)=1+x$ Prove that $f(1) \leq \frac{4}{3}$.

Given solution (which I didn't understand):

$1+x$ is never $0$, so $1+f(x)$ is never $0$. As $f(0)=1$, it is always positive. Hence also $f''(x)$ is always positive. As $f'(0)=0$, this gives $f'(x)>0, \forall x>0$. Hence in particular $1+f(x) \geq 2 \implies f''(x) \leq \frac{1+x}{2}$. Integrating gives $f'(x) \leq f'(0) + \frac{x}{2}+\frac{x^2}{4} = \frac{x}{2}+\frac{x^2}{4}$. Integrating again gives $f(x) \leq f(0)+\frac{x^2}{4}+\frac{x^3}{12}$.
  Hence $f(1) \leq 1+1/4+1/12=4/3$.

I failed to see why $f(x)$ must always be positive. I can understand the rest of the solution apart from this. 
I am also unable to understand the situation by considering some functions as examples, as I'm unable to think of any elementary function which satisfies the given conditions.

Since $f''(x)$ is always positive, $f'(x)$ must also always be positive since $f'(0) = 0$. If not, then take some $a$ such that $f'(a) \le 0$. By the Mean Value Theorem, we obtain $\frac{f'(a)-f'(0)}{a-0} = f''(c)>0$ for some $c$, a contradiction as $\frac{f'(a)-f'(0)}{a-0} = \frac{f'(a)}{a} \le 0$. Repeat the same argument on $f(x)$, to get $f(x)>1$ for $x>0$.
