Integral inequality with non-increasing function and integral bound

Question

Let $f,g$ be two integrable functions on $[a,b]$ with $f$ non-increasing and $0\le g\le 1$. Let $$\lambda=\int_a^b g(x)dx$$ Prove that $$\int_{b-\lambda}^b f(x)dx\le \int_a^b f(x)g(x)dx$$

I've tried to transform this inequality into $\int_{b-\lambda}^b f(x)(1-g(x))dx\le \int_a^{b-\lambda} f(x)g(x)dx$, and estimate two sides with $f$ non-increasing condition, but don't know how to deal with $\lambda$. I also tried using intermediate value theorem to write $\lambda=(b-a)g(t)$ for some $t\in [a,b]$, but still failed to proceed. Thanks for any help.

Answer 1

Without loss of generality, let $a = 0$. Define \begin{align*} h(b) = \int_0^b f(x)g(x) dx - \int_{b - \lambda_b}^b f(x) dx \end{align*} where $\lambda_b = \int_{0}^{b}g(x) dx \ge 0$. Hence we have \begin{align*} h'(b) = f(b)g(b) - [f(b) - f(b - \lambda_b)(1-g(b))] = [f(b - \lambda_b) - f(b)][1 - g(b)] \end{align*} Now $f$ is non-increasing $\implies f(b - \lambda_b) \ge f(b)$ and $g(b) \in [0, 1] \implies 1 - g(b) \ge 0$, hence $h'(b) \ge 0$ and $h$ is a non-decreasing function. This implies that $h(b) \ge h(0)$ for all $b \ge 0$, and it's easy to check that $h(0) = 0$, and the result follows.