$\int_0^s \log f(t)dt \le \int_0^s \log g(t)dt$ for every $s\in (0,1)$ implies that $\int_0^1 \log (f(t)+1)dt \le \int_0^1 \log (g(t)+1)dt$?

Question

Assume that $f,g$ are decreasing positive functions on $(0,1)$. If $\int_0^s \log f(t)dt \le \int_0^s \log g(t)dt$ for every $s\in (0,1)$, then is it true that $\int_0^1 \log (f(t)+1)dt \le \int_0^1 \log (g(t)+1)dt$?

I guess that we can use something similar to (1) of https://hrcak.srce.hr/file/149449
Since $p(x)= \log(e^x +1)$. Then, $p(log(f)) =log(f+1)$. Since $p$ is a continuous convex function. Then, the assertion follows from (1) of https://hrcak.srce.hr/file/149449

Answer 1

It is the integral version of the Hardy-Littlewood-Polya inequality.

see [Chapter 1, Theorem D.2] of A. Marshall, I. Olkin, B. Arnold, Inequalities: theory of majorization and its applications, second edition, Springer series in statistics, Springer, New York, 2011.

See also J. Ryff, On muirhead's theorem Pacific J. Math. 21 (1967), 567--576.

[Theorem 12.15] J. Pecaric, F. proschan, Y. Tong, Convex functions, partial orderings and statistical applica- tions, Academic Press, 1992.