Proving an Integral inequality from a given integral inequality

Question

Problem: Let $f$ and $g$ be continuous, non-negative function on $[0, 1]$, with

$$\int_{0}^{1}e^{-f(x)}dx \geq \int_{0}^{1}e^{-g(x)}dx. $$

Prove that,

$$\int_{0}^{1}g(x)e^{-f(x)}dx \geq \int_{0}^{1}f(x)e^{-g(x)}dx. $$

This problem is an equivalent form of a problem from the book "More Calculus of a variable" by Peter Mercer. In the original problem there was $xf(x)$ and $xg(x)$ in place of $f(x)$ and $g(x)$. I couldn't make much progress. I tried contradiction and method which led to conlclude that its enough to show,

$$e^{g(x)}-e^{f(x)}+f(x)e^{f(x)}-g(x)e^{g(x)} < 0.$$ But sadly this is not true in general.

I also tried integration by parts to seperate $\displaystyle\int^{1}_{0}e^{-g(x)}\,dx$ from $\displaystyle\int^{1}_{0}f(x)e^{-g(x)}\,dx$ to use the given inequality. But I didn't happen.

Please help me. I am in high school so please don't use any very advanced inequality.

Answer 1

From the original inequality: $$\int_0^1 e^{-f(x)}dx\ge\int_0^1e^{-g(x)}dx \rightarrow f(x)\le g(x)$$ Normally you would want to prove this specifically but if you are in highschool we will just take it as intuitive. Also, we have the statement: $$e^{-f(x)} \ge e^{-g(x)}$$ Using this: $$g(x)e^{-f(x)} \ge f(x)e^{-f(x)}$$ But also: $$f(x)e^{-f(x)} \ge f(x)e^{-g(x)}$$ Therefore: $$g(x)e^{-f(x)} \ge f(x)e^{-f(x)}\ge f(x)e^{-g(x)}$$ Drop the middle term then integrate: $$g(x)e^{-f(x)} \ge f(x)e^{-g(x)} \rightarrow \int_0^1 g(x)e^{-f(x)} dx \ge \int_0^1f(x)e^{-g(x)}dx$$

Answer 2

Your conclusion does not hold. As an example, take $$ f(x) = x \, , \quad g(x) = 1 - \log(e-1) \, . $$ Then $$ \int_0^1 e^{-f(x)} \, dx = \int_0^1 e^{-g(x)} \, dx = 1 - \frac 1e \, , $$ but $$ \int_0^1 g(x) e^{-f(x)} \, dx = (1 - \log(e-1) ) (1 - \frac 1e) \approx 0.29 \\ < \int_0^1 f(x)e^{-g(x)} \, dx = \frac 12 (1 - \frac 1e) \approx 0.31 \, . $$