Prove $f (1)<\frac32$ given $f (0)=0$, $f'(0) = 1$, $f(x)+1=\frac1{f''(x)}$ Let $f: \mathbb {R}\rightarrow[0,+\infty)$ be a two times differentiable function with $f (0)=0$, $f'(0) = 1$, and $f(x)+1=\dfrac {1}{f''(x)}$ for $x\in [0,1]$. Show that  $f (1)<\dfrac {3}{2} $ 
So far I have tried to apply the mean value theorem and find out that $f (1) = f'(c) $, for some $c\in(0,1) $
Applying the same to the second derivative, I got
$$f'(1) - 1 = f''(k) $$
$$f'(1) = \dfrac {1}{1+f (k)} + 1$$
For some $k\in(0,1) $
Also finded that $f''(0) = 1$. , and $f''(x) $ is never $0$.
But I can't advance from here. Any hints?
Thanks!
 A: First,
$$
f'' = \frac{1}{1+f} \Rightarrow f'f'' = \frac{f'}{1+f}.
$$
Then
$$
 \frac{d}{dx}\left(\frac{1}{2}{f'}^2\right) = \frac{d}{dx}\left(\ln(|1+f|)\right).
$$
This implies below. ($f(0)=0$ and $f'(0) = 1$ gives constant precisely.)
$$
\frac{1}{2}{f'}^2 = \ln(|1+f|) + \frac{1}{2}
$$
Thus  $|f'| = \sqrt{2\ln(|1+f|) + 1}$ .

Claim: $f>0$ and $f'>0$ on $(0,1]$. - (*)
Argument)
Note that this argument is not mathematically perfect. If you want analytical details, please see bottom of the answer.
From $f'' = \frac{1}{1+f}$ , $f''(0)> 0$, thus $f'$ increases near $x=0.$ This implies $f'(x)>1>0\,\,$for $x$ slightly greater than $0$. This implies $f$ increases near $x=0$. Thus $f(x)>0\,\,$for $x$ slightly greater than $0$. Then $f''(x) = \frac{1}{1+f(x)} > 0$ .. and so on.

Then  $f' = \sqrt{2\ln(1+f) + 1}$ on $x \in [0,1]$.
Since  $0 \leq f' = \sqrt{2\ln(1+f) + 1} \leq \sqrt{2f + 1}$ (Note second equality only holds when $x=0$),
$$
0\leq \frac{f'}{\sqrt{2f + 1}}  <1 \quad(x\in(0,1]) \Rightarrow 0\leq \frac{d}{dx}{\sqrt{2f + 1}}  <1 \quad(x\in(0,1])
$$ 
Thus we get
$$
0 \leq \int_0^{x} \frac{d}{dt}\sqrt{2f(t) + 1}\,\, dt < x \,\,\Rightarrow\,\, 0 \leq \sqrt{2f(x) + 1} - 1  <x \,\,(x\in(0,1]),\\ \text{then } \sqrt{2f(1) + 1} - 1  <1\,.
$$ 
Equivalently
$$
f(1)< \frac{3}{2}.
$$

Proof for $(*)$: First, I will show $f'>0$ on $(0,1]$.
Assume to the contrary, let $\exists x \in (0,1]$ satisfies $f'(x) \leq 0$.
Then $A \equiv \{x\in(0,1] : f'(x) \leq 0\}$ is nonempty. Let $x_0 = \inf(A)$.
From continuity of $f'$, $\,f'(x_0) \leq 0$. Then by mean value theorem(MVT) on $f'$ between $0$ and $x_0$, there exists $c_1 \in (0,x_0)$ such that $f''(c_1) <0$. Then relation above gives $f(c_1) < 0$. 
Again, by MVT between $0$ and $c_1$, there exists $c_2 \in (0,c_1)$ such that $f'(c_2) < 0$. ($\Rightarrow c_2 \in A$)
This contradicts with $x_0 = \inf(A)$.
Now '$\,f>0\,$ on $(0,1]\,$' follows from $f(x) = \displaystyle{\int_0^x{f'(t)}} \,\,>0\,\,$ ($x\in(0,1]$).
A: different way to look at it; the Taylor series begins
$$  x + \frac{x^2}{2} - \frac{x^3}{6} +\frac{x^4}{24}   +\frac{x^5}{120} - \frac{15x^6}{720}  +\frac{79x^7}{5040} - \frac{223x^8}{8!} - \frac{863x^9}{9!} + \frac{19841x^{10}}{10!} - \frac{177025x^{11}}{11!} $$
and the value of the function at $x=1$ is the sum of the series givenn by erasing all the $x$'s. So, a matter of showing that the numerators $a_n$ in $\frac{a_n x^n}{n!}$ grow more slowly than $n! \; , \;$ and that the partial sum to that point is below $3/2$

just something to think about
A: We have $f(0) = 0$, $f'(0) = 1$, and
$$ \tag{*}
 f''(x) (1+ f(x)) = 1 \quad \text{for } x \ge 0 \, ,
$$
so that $f''(0) = 1$. Now we can draw the following conclusions:


*

*$f''(x) > 0$ for $x \ge 0$: $(*)$ implies that $f''$ is continuous and has no zeros, so that it has the same sign everywhere (intermediate value property).

*$f'$ is strictly  increasing, $f'(x) > f'(0) = 1$ for $x > 0$.

*$f$ is strictly increasing. Using $(*)$ again it follows that

*$f''$ is strictly decreasing, $f''(x) < f''(0) = 1$ for $x > 0$.


Integrating $f''(x) < 1$ twice (or using Taylor's theorem) gives
$$
 f(x) < f(0) + x f'(0) + \frac 12 x^2 = x + \frac 12 x^2
$$
for $x > 0$, and in particular $f(1) < 3/2$.

With little more effort we can obtain a slightly better bound: We have $f'(x) > 1$, so that $f(x) > x$ and therefore
$$
 f''(x) < \frac{1}{1+x}
$$
for $x > 0$. Integrating this twice gives
$$
f(x) < x + \int_0^x \log(1+t) \, dt = (1+x) \log(1+x)
$$
and in particular $f(1) < 2 \log 2 \approx 1.386$.
A: You may like this. Note
$$ f(1)=f(0)+f'(0)\cdot1+\frac12f''(c)\cdot 1^2=1+\frac12f''(c) $$
where $c\in(0,1)$. From 
$$f''(x)(1+f(x))=1,\tag{1}$$ 
one has $f''(0)=1$. It is easy to see that $f''(x)>0$ for $x\in[0,1]$. Since otherwise, one would have some $x_0\in[0,1]$ such that $f''(x_0)=0$ which is against (1). So $f'(x)>0$ is strictly increasing and hence $f(x)>f(0)=0$ for $x>0$. So$f(c)>0$ and hence
$$ f(1)=1+\frac12f''(c)=1+\frac12\frac1{1+f(c)}<1+\frac12=\frac32. $$
