An integral inequality revisited

Question

Assume that $\Phi, \Psi$ are positive increasing functions and $g$ positive non-increasing so that $$\int_0^1 \Phi\left(\frac{g(t)}{t}\right)dt = \int_0^1 \Phi\left(\frac{1}{t}\right)dt=1.$$

Then it seems to me that $$\int_0^1 \Phi\left(\frac{g(t)}{t}\right)\Psi(t)dt\le \int_0^1 \Phi\left(\frac{1}{t}\right)\Psi(t)dt?$$

Answer 1

Choose $a\in [0,1]$ such that $g(t)\geqslant 1$ for $t\leqslant a$ and $g(t)\leqslant 1$ for $t\geqslant a$. Then $(\Phi(g(t)/t)-\Phi(1/t))(\Psi(t)—\Psi(a))\leqslant 0$ for all $t\in [0,1] $. Integrate it.