Prove or disprove that $\int_a^bf(x)g(x)dx\geqslant g(b)\frac{\int_a^bf(x)dx}{b-a}$.

Question

Let $f:[a,b]\to\mathbb{R}$ and $g:[a,b]\to\mathbb{R}$ be continuous, positive-valued functions on interval $[a,b]$ and let $g$ be decreasing. Then prove or disprove the following inequality $$\int_a^bf(x)g(x)dx\geqslant g(b)\frac{\int_a^bf(x)dx}{b-a}.$$ I intuitively think that it is true but I can't prove it. I've just came up with this and I am not certain if it's true for sure. I don't know if assumptions are too weak or if they can be loosened up.

Answer 1

Counterexample: $a:=1, b:=1.5$. Let $f(x) = 1, g(x) = 1/x$. Then both continuous, positive, $g$ strictly decreasing on $[1, 1.5]$.

Then LHS $=\ln 1.5 < 0.41 $ whereas RHS $=2/3$.

Edit (answer to comment). If the constraint $b-a\geq 1$ is added, then the modified statement is true because

$$\int\limits_{a}^{b}{f(x)g(x)}\,\mathrm{d}x \geq g(b) \int\limits_{a}^{b}{f(x)}\,\mathrm{d}x \geq g(b)\dfrac{\int\limits_{a}^{b}{f(x)}\,\mathrm{d}x}{b-a}.$$

Answer 2

This is true if you remove the $b-a$ from the right side.

As stated it is false. If f and g are constants, the left side is $(b-a)fg$ and the right side is $fg$. If the $b-a$ is removed from the right side, it becomes $fg$ snd the two sides are equal.