Assume that $f,g$ are decreasing positive functions on $(0,1)$. If $\int_0^1 \log f(t)dt \le \int_0^1 \log g(t)dt$, then is it true that $\int_0^1 \log (f(t)+1)dt \le \int_0^1 \log (g(t)+1)dt$?
2026-03-25 08:13:37.1774426417
$\int_0^1 \log f(t)dt \le \int_0^1 \log g(t)dt$ implies $\int_0^1 \log (f(t)+1)dt \le \int_0^1 \log (g(t)+1)dt$?
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No. Take $g(t)$ as $e-1$ and $f(t)$ as $e^{(1-t^2)^2}$.