I'm trying to prove that, given some finite numbers $a$ and $b$ ($a < b$) and a non-negative,continuous,smooth function $f(k,x)$, the following inequality:
$$\int_a^b e^{-f(k_1,x)}\;\mathrm{d}x > \int_a^b e^{-f(k_2,x)}\mathrm{d}x$$
can be simplified to
$$\int_a^b f(k_1,x)\; \mathrm{d}x < \int_a^b f(k_2,x) \; \mathrm{d}x$$
For any some $k_1,k_2 \in \mathbb{R}$. Does this hold? If yes, could you give me a hint on how to prove it.
The claim is false. On the interval $[a,b]=[0,2]$, consider $g(x)=\frac25$ and $h(x) = \lfloor x\rfloor$. Then $$ \int_0^2 g(x)\,dx = \frac45 < 1 = \int_0^2 h(x)\,dx $$ and yet $$ \int_0^2 e^{-g(x)} = 2e^{-2/5} \approx 1.34 < 1.37 \approx 1+e^{-1} = \int_0^2 e^{-h(x)} \,dx. $$ It just remains to choose a function $f(k,x)$ such that $f(0,x) = g(x)$ and $f(1,x) = h(x)$, say.