Let $f(x)$ be defined and continuous and derivable for $x>-1$, $f(0)=1$, $f’(0)=0$ and $$f''(x) = \frac {1+x}{1+f(x)}.$$ Prove that $f’(x)$ is concave up for all $x>-1$.
My attempt: I tried to integrate by multiplying both sides by $dy/dx$ but could not proceed further.
On
Suppose to the contrary that $f^{'}$ is convex on $(-1,+\infty)$. From $f^{''}=\frac{1+x}{1+f}$, we know $f(x)\neq-1$ for any $x\in(-1,+\infty)$; as otherwise $f^{''}(x)$ does not exist. Therefore $f^{'''}$ exists on $(-1,+\infty)$. Then the convexity of $f^{'}$ implies that $f^{'''}\ge0$. So for all $x>0$, $$f^{''}(x)\ge f^{''}(0)=\frac{1+0}{1+f(0)}=\frac12.~~~~~~~~~~~~~~(1)$$ It follows that $f(x)\ge\frac14x^2+1$ for all $x>0$. But this gives us that $$f^{''}(x)\le\frac{1+x}{2+\frac14x^2}\rightarrow0 ~~\mathrm{when} ~~x\rightarrow+\infty,$$ which contradicts $(1)$.
To see this, we first claim that $f(x)>-1$ for any $x\in(-1,0)$. We have shown that for $x>−1, f(x)\neq-1$. Suppose that $f(x_0)<−1$ for some $x_0 \in (−1,0)$. Since $f$ is continious and $f(0)>-1$, by Intermediate Value Theorem, there exists some $\xi\in(x_0,0)$ such that $f(\xi)=−1$. This gives us a contradiction.
Now for any $x\in(-1,0)$, as $f(x)>-1$, by $f^{''}(x)=\frac{1+x}{1+f(x)}$ we have that $f^{''}(x)>0$. So for all $x\in(-1,0)$, $f^{'}(x)\le f^{'}(0)=0$. Thus for $x\in(-1,0)$, $$f^{'''}(x)=\frac{1+f(x)-(1+x)f^{'}(x)}{(1+f(x))^2}\ge\frac1{1+f(x)}>0.$$ This implies $f^{'}$ is convex on $(-1,0)$.
Let $h(x)=f'(x)$, $x\in(-1,+\infty)$ and $$ f''(x)=\frac{1+x}{1+f(x)},\tag 1 $$ then $$ h''(x)=f'''(x)=\frac{d}{dx}\left(\frac{1+x}{1+f(x)}\right)=\frac{1+f(x)-(1+x)f'(x)}{(1+f(x))^2}. $$ Set $$ g(x):=1+f(x)-(1+x)f'(x)\textrm{, }x>-1, $$ then $$ g'(x)=f'(x)-f'(x)-(1+x)f''(x)=-\frac{(1+x)^2}{1+f(x)}. $$ But $f(x)+1>0$, $\forall x\in(-1,+\infty)$. Otherwise if exists $x_0>-1$ such $f(x_0)+1<0$, then we define the function $P(x)=f(x)+1$, $x>-1$. Then $P(x_0)<0$ and $P(0)=f(0)+1=2>0$. Hence from Bolzano theorem exists $\rho\in(x_0,0)$ (resp. $\rho\in(0,x_0))$ such $P(\rho)=0$ and hence $f(\rho)=-1$. But from (1) we get $f''(\rho)=\infty$, if $\rho\neq-1$. Hence $\rho=-1$, which is also contradiction.
Hence $f(x)>-1$, $\forall x\in(-1,+\infty)$ and therefore $$ g'(x)<0\textrm{, }\forall x\in(-1,+\infty). $$ From this last result $g(x)$ is dicreasing in $(-1,+\infty)$. Also we conclude that
$$ f''(x)>0\textrm{, }\forall x\in (-1,+\infty). $$ Hence $f'(x)$ is increasing. Hence for $x>0$ we get $f'(x)>f'(0)=0$ and if $-1<x<0$, then $f(-1)<f'(x)< f'(0)=0$. Hence $$ f(x)\geq f(0)=1\textrm{, }\forall x\in (-1,+\infty) $$
If $g(x)=0$ has no solution in $(-1,+\infty)$, then from $g(0)=1+1=2>0$, we get $g(x)>0$ and hence $h''(x)=f'''(x)>0$, for all $x>-1$. Hence for $x \geq 0$, $f''(x)\geq f''(0)=\frac{1+0}{1+f(0)}=\frac{1}{2}$. Hence $f'(x)-f'(0)\geq\frac{x}{2}\Rightarrow f'(x)\geq \frac{x}{2}\Rightarrow f(x)\geq\frac{x^2}{4}+1$. Setting this in (1) we get as Eric Yau did $$ \frac{1}{2}\leq f''(x)\leq \frac{x+1}{\frac{1}{4}x^2+2}\rightarrow 0\textrm{, }x\rightarrow \infty, $$ which is contradiction.
Hence exists $x_1$ such that $g(x_1)=0$ and for $x>x_1$, $g(x)<g(x_1)=0$. Also for $-1+\epsilon<x<x_1$, $\epsilon>0$, we have $g(-1+\epsilon)>g(x)>0$.
Hence $f'''(x)<0$, for $x>x_1$ and $f'''(x)>0$, for $-1+\epsilon<x<x_1$ and $0<f''(x)=h'(x)\leq h'(x_1)$, for all $x>-1+\epsilon$. Hence $f''(x)$ is positive and bounded above by $f''(x_1)$.
If where more than one roots (say $-1<x_1<x_2$) of $g(x)=0$ in $(-1,+\infty)$, then exist $\xi\in(x_1,x_2)$ such that $g'(\xi)=0$. But we know that $g'(x)<0$ in $(-1,+\infty)$, and this is not permitable.
Also $x_1>0$ because if $-1<x_1 \leq 0\Rightarrow g(0)\leq g(x_1)<g(-1)\Rightarrow 2<g(x_1)=0$ (not true).
Hence $f'''(0)=\frac{2}{4}=\frac{1}{2}>0$ and we conclude that exists a $x_1>0$ such that
$$ f'''(x)>0\textrm{, if }x\in(-1,x_1) $$
and $$ f'''(x)<0\textrm{, if }x\in(x_1,+\infty). $$ Note. $x_1\approx 1.4588\pm 10^{-5}$.