Assume that all of the following functions and expressions are defined in such a way that the following make sense.
Let $ f:[0,1] \to \mathbb R $ and set $ g(t):= - \log f(t)$ and $ h(t):= f(t)^{1/n}$, where $ n \in \mathbb N.$ Suppose that the following differential inequality holds:
$$ g'' (t) \geq \frac{1}{n} g'(t)^{2} $$
Can we obtain from the above differential inequality an inequality of the following form?
$$ h(t) \geq (1-t) h(0) + t h(1), \quad t \in [0,1]$$
From $$g(t)= - \log f(t)$$ we learn that $f(t)>0$ and that
$$g'(t)= - \frac {f'(t)}{f(t)}$$ $$g''(t)= - \frac{f''f-(f')^2}{f^2}\ge \frac{\left(- \frac {f'(t)}{f(t)}\right)^2}{n}$$
Rewrite the inequality for $h$ in terms of $f$ to get concavity definition for $f^{1/n}(t)$
$$f^{1/n}(t) \geq (1-t) f^{1/n}(0) + t f^{1/n}(1), \quad t \in [0,1]$$ which is holds since $$\begin{align} \frac{d}{dt}f^{1/n}(t)&=\\ &=\frac{f^{\frac{1}{n}-2} \left(n f''f-(n-1) (f')^2\right)}{n^2}\\ &=f^{\frac{1}{n}} \left( \frac{f''f-(f')^2}{f^2}+ \frac{(f')^2}{nf^2}\right) <0\end{align}$$