Proving an Integral inequality from a given integral inequality 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 &quot;More Calculus of a variable&quot; 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,
$$e^{g(x)}-e^{f(x)}+f(x)e^{f(x)}-g(x)e^{g(x)} &lt; 0.$$ 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.
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
$$
\int_0^1 g(x) e^{-f(x)} \, dx = (1 - \log(e-1) ) (1 - \frac 1e) \approx 0.29 \\
 &lt; \int_0^1 f(x)e^{-g(x)} \, dx = \frac 12 (1 - \frac 1e) \approx 0.31 \, .
$$
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:
$$g(x)e^{-f(x)}  \ge f(x)e^{-g(x)} \rightarrow \int_0^1 g(x)e^{-f(x)} dx  \ge \int_0^1f(x)e^{-g(x)}dx$$