If $f:\mathbb{R}_+\to\mathbb{R}_+$ is strictly concave and strictly increasing with (...), is it true that $f''((f')^{-1}(x))\cdot(f')^{-1}(x)>-x$? Is this true? If $f:\mathbb{R}_+\to\mathbb{R}_+$ is strictly concave and strictly increasing with $f(0)=0$, $\lim_{x\to0^+}f'(x)=\infty$ and $\lim_{x\to\infty}f'(x)=0$, then $$f''((f')^{-1}(x))\cdot (f')^{-1}(x)>-x$$ for all $x>1$. 
Here $\mathbb{R}_+$ is the set of nonnegative real numbers.
Here are some simplifications. Since $f'((f')^{-1}(x))=x$ differentiation with respect to $x$ gives $f''((f')^{-1}(x))\cdot\frac{d}{dx}(f')^{-1}(x)=1$. Consequently, $$f''((f')^{-1}(x))\cdot (f')^{-1}(x)=\frac{1}{\frac{d}{dx}(f')^{-1}(x)}\cdot(f')^{-1}(x)=\frac{d}{dx}\log{(f')^{-1}(x)}.$$ The graph of $f$ is represented below.

 A: This is false. Put $g(x) = f'(x)$. Then the question becomes: when $g : \mathbb R^+ \to \mathbb R^+$ is smooth, integrable at $0$ and strictly decreasing with $\lim_{x \to 0^+} g(x) = +\infty$ and $\lim_{x \to +\infty} g(x) = 0$, is it true that
$$g'(x) \cdot x > - g(x)$$
whenever $g(x) > 1$?
Put $h(x) = \log g(x)$, then the question becomes: when $h : \mathbb R^+ \to \mathbb R$ is smooth, $e^{h(x)}$ integrable at $0$ and $h(x)$ strictly decreasing with $\lim_{x \to 0^+} h(x) = +\infty$ and $\lim_{x \to +\infty} h(x) = -\infty$, is it true that
$$h'(x) > -\frac1x$$
whenever $h(x) > 0$?

We can now draw the graph of a counterexample: take $h(x)$ to grow not too fast at the origin (like $-0.5\log x$); before it dives below $0$ let it decrease very very fast, then continue the graph by letting $h(x)$ decrease to $-\infty$ in any way you like.
Because $h(x)$ decreases very fast at some point $x_0$ with $h(x_0) > 0$, we will have
$$h'(x_0) < -\frac1{x_0} \,.$$
More explicitly, take a smooth decreasing $h$ that satisfies
$$
h(x) = \begin{cases} -0.5 \log x & x \leq 0.001 \\
-1000 (x-2) & 1.999 \leq x \leq 2.001 \\
-x & 3 \leq x
\end{cases}$$
and $x_0 = 1.999$. We have $h'(x_0) \approx -500$ and $-1/x_0 \approx -0.5$.
Finally, put $f(x) = \int_0^x e^{h(t)} dt$.
