Prove this functional inequality Given $f: R \to (0,\infty]$, twice differentiable with $f(0) = 0, f'(0) = 1, 1 + f = 1/f''$
Prove that $f(1) < 3/2$
I found some useless (I believe) facts about $f$, but nothing that gives me a light to answer it.

*

*$f''(x) > 0$

*$f'(x) \ge 1$

*$e^{((f')^2-1)/2} = f+1$ is another way to write the differential equation

....
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As you've noticed, this can be written as
$$\frac12(f'^2-1)=\ln(1+f).$$ If we want a parametric equation, we might choose $f'=t$ as a parameter, and then, we have (denoting the derivative with respect to $t$ by a point)
$$f(t)=e^{(t^2-1)/2}-1$$ and $$\dot x=\dot f/t=e^{(t^2-1)/2},$$ and since $x=0$ corresponds to $t=f'(0)=1$, $$x(t)=\int^t_1 e^{(u^2-1)/2}\,du.$$ The latter may not be expressible by "elementary" functions, but there are a lot of rather common special functions one can use for that (Dawson's integral, error function or incomplete gamma function at some imaginary argument). So we can numerically solve for the value of $t$ that corresponds to $x=1$, finding $t\approx1.6517245235869685402116665754535807279$, and there, $f\approx1.3728623070052740075133959835820792705$.
