<s>[INST] Integral inequality $\int_0^1 f(x)\,dx \ge 4 \int_0^{1\over2} f(x)\,dx$? If $f: [0, 1] \to \mathbb{R}$ is a convex and integrable function with $f(0) = 0$, does it necessarily follow that$$\int_0^1 f(x)\,dx \ge 4 \int_0^{1\over2} f(x)\,dx?$$
 [/INST] For $0\le x\le \frac{1}{2}$, by convexity one has $$f\left(\frac{1}{2}\right)\le \frac{1}{2}\left[f\left(\frac{1}{2} + x\right) + f\left(\frac{1}{2} -x\right)\right].$$ Also, $$\int_0^1f(x)dx=\int_0^{\frac{1}{2}}f(x)dx +\int_{\frac{1}{2}}^1f(x)dx=\int_0^{\frac{1}{2}}f(x)dx +\int_0^{\frac{1}{2}}f\left(x+\frac{1}{2}\right)dx.$$  From the first inequality and the above, one has $$\int_0^1f(x)dx\ge\int_0^{\frac{1}{2}}f\left(x\right)dx + 2f\left(\frac{1}{2}\right)\int_0^{\frac{1}{2}}dx - \int_0^{\frac{1}{2}}f\left(\frac{1}{2}-x\right)dx.$$ The last and the first integral in the above inequality cancel. Finally note that by convexity and the condition $f(0)=0$ $$\int_0^{\frac{1}{2}}f(x)dx\le \frac{1}{2}\frac{1}{2}f(\frac{1}{2}).$$  The required inequality follows from above.
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