Let $f:[a,b]\rightarrow \mathbb{R}$ be a function. $f'_{+}(t)= \lim_{t'\rightarrow t^{+}}\frac {f(t')-f(t)}{t'-t}$.
Let $f:[0,1]\rightarrow \mathbb{R}$ be a concave function. Then this means $\frac{f(t)-f(0)}{t}\geq \frac{f(t')-f(0)}{t'}$ for $0<t<t'$. Then as here slope is decreasing function, $f'_{+}(t)$ is equal to infimum of $\frac{f(t)-f(0)}{t}$ over all $t>0$ SO why is the following true:
true that $f'_{+}(t)\geq \frac{f(s)-f(t)}{s-t}$?
$f$ concave means $f(x+t(y-x))\geq f(x)+t(f(y)-f(x))$ So $f(y)-f(x)\leq$ $\lim_{t\rightarrow 0+} \frac{f(x+t(y-x))-f(x)}{t}$
Write $h=t(y-x)$. Then, $t\rightarrow 0$ iff $h\rightarrow 0$. So,
$\lim_{t\rightarrow 0^{+}}\frac{f(x+t(y-x))-f(x)}{t}=\lim_{h\rightarrow 0^{+}}\frac{(f(x+h)-f(x))(y-x)}{h}=f'_{+}(x)(y-x)$ So, $f'_{+}(x)(y-x)\geq f(y)-f(x)$