How prove this integral inequality $\int_{0}^{s}f(x)\,dx\le\int_{s}^{1}f(x)\,dx\le\dfrac{s}{1-s}\int_{0}^{s}f(x)\,dx$
Without loss of generality, we may assume that $\int_0^1f(x)dx=1$.
Define $F(x)=\int_0^x f(t)dt$. By definition, $F(0)=0$ and $F(1)=1$. Moreover, since $F'=f$ is positive and increasing, $F$ is increasing and convex. Therefore, by Jensen's inequality,
$$F(s)=F\big(\int_0^1xf(x)dx\big)\le \int_0^1F(x)f(x)dx=\int_0^1F(x)F'(x)dx=\frac{1}{2}.\tag{1}$$
It follows that
$$\int_0^sf(x)dx=F(s)\le 1-F(s)=\int_s^1f(x)dx.\tag{2}$$
By the convexity of $F$, when $0\le t\le 1$,
$$F(ts)\le tF(s)+(1-t)F(0)=tF(s)\tag{3}$$
and
$$F(ts+1-t)\le tF(s)+(1-t)F(1)=tF(s)+1-t.\tag{4}$$
Due to $(3)$ and $(4)$, we have
$$\int_0^s F(x)dx=s\int_0^1F(ts)dt\le\frac{s}{2}F(s)\tag{5}$$
and
$$\int_s^1 F(x)dx=(1-s)\int_0^1F(ts+1-t)dt\le\frac{1-s}{2}(F(s)+1).\tag{6}$$
$(5)+(6)$ implies that
$$\frac{1}{2}(F(s)+1-s)\ge\int_0^1 F(x)dx= xF(x)\big|_0^1-\int_0^1xf(x)dx =1-s.\tag{7}$$
It follows that
$$\int_s^1f(x)dx=1-F(s)\le\frac{s}{1-s}F(s)=\frac{s}{1-s}\int_0^sf(x)dx.\tag{8}$$