$D>0, L>0$ are constant. $f:[0, +\infty) \rightarrow \mathbb R$ is non-negative continuous function. $f(0)=0$ and $D=\int_0^{+\infty} f(x) dx$. In fact, when $x>0$ is large enough (than $L$), there is $f(x)=0$. If $$ f(x)\ge 2x-\frac{L}{D} \int_0^x f(t)dt ~~~~~~\forall x\in [0,\frac{L}{2}] \tag{1} $$ Then, how to show $$ \int_0^{L/2} f(x)dx \ge CL^2 \tag{2} $$ where $C$ is a positive constant. The reason of $(2)$ is that the problem has actual background. In fact, I want to calculate the optimal $C$.
What I try: Integrate $(1)$, I have $$ \int_0^{L/2} f(x)dx\ge \frac{L^2}{4}-\frac{L}{D}\int_0^{L/2}\int_0^x f(t)dt dx $$ After exchange the order of integrate, I have $$ \int_0^{L/2} f(x)dx\ge \frac{L^2}{4}-\frac{L}{D}\int_0^{L/2} f(t)(\frac{L}{2}-t)dt \tag{3} $$