# Is $\int_0^t \left[f(x)-g(x) + l\right]\mathrm dx > \int_t^1 \left[f(x-t)-f(x)\right] \mathrm dx$?

Consider $f,g$ where $f(x) > g(x) \ge 0 \ \forall x \in (0,1)$ and $f(0) = g(0)$, $f(1) = g(1)$ . Is the following inequality true?

$$\int_0^t \left[f(x)-g(x) + l\right]\mathrm dx > \int_t^1 \left[f(x-t)-f(x)\right] \mathrm dx$$

for any $l > 0, 0<t<1$

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The inequality is not true. For a counter-example (for sure not the easiest) consider $$f(x) = 1 -\frac{x}{2},$$ $$g(x)=f(x) - x(1-x)$$ and $l=\frac{1}{24}$ and $t=\frac{1}{2}$. Then the left hand side is $\frac{5}{48}$ and the right hand side is $\frac{1}{8} = \frac{6}{48}$.
Just nitpicking, $f(0)\neq 0$ in your definition, and $f(1)\neq 1$. By that $g(0)\neq 0$ and $g(1)\neq 1$ as well. –  Asaf Karagila May 7 '11 at 16:36
Fabian: The question asks "Given $f,g$ such that ... is the following inequality true?". You specify a counterexample, but the conditions in the questions do not hold for it. –  Asaf Karagila May 7 '11 at 16:44
@Asaf: Actually Fabian's counterexample works. The question only requires $f(0) = g(0)$ and $f(1) = g(1)$ . In his construction $f(0) = 1 = g(0)$ and $f(1) = 0.5 = g(1)$. –  snel May 7 '11 at 16:51