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$$\int_0^1 f(x)\,dx \ge 4 \int_0^{1\over2} f(x)\,dx?$$
 A: Yes, this is true: \begin{align}\frac12\int_{0}^1f(x)dx&\overset{(1)}=\int_{0}^1\left(\frac12f(x)+\frac12f(0)\right)dx\\&\overset{(2)}\ge \int_{0}^1f\left(\frac12x+\frac120\right)dx=^{y=\frac12x}\\&=\int_{0}^{1/2}f(y)2\ dy\end{align} where $(1)$ holds by $f(0)=0$ and $(2)$ by the convexity of $f$.
From this solution, we can infer the more general result 
$$\int_0^1 f(x)dx\ge (1/λ)^2 \int_0^λ f(x)dx$$ for any $λ\in (0,1)$.
A: The function
$$g(x):=f(x)-2x f\left({1\over2}\right)$$
is convex and satisfies $g(0)=g\bigl({1\over2}\bigr)=0$. It follows that $g(x)\leq0$ when $0\leq x\leq{1\over2}$, and $g(x)\geq0$ when ${1\over2}\leq x\leq1$. This implies
$$\int_{1/2}^1 g(x)\>dx\geq 3\int_0^{1/2} g(x)\>dx\ ,$$
hence
$$\int_0^1 g(x)\>dx\geq 4\int_0^{1/2} g(x)\>dx\ .\tag{1}$$
Since for linear functions $g:\>x\mapsto cx$ we have equality sign in $(1)$ it follows that $(1)$ holds also for the function $f(x)=g(x)+2x f\bigl({1\over2}\bigr)$. 
A: Let $L : y=ax$ be the line passing through the points $A =(0,0) $ and $B=(0.5 , f(0.5 ) $ and let $C =(0.5 ,0)$ , $D =(1,a) , E = (1,0)$
Then $$\int_{0}^{0.5} f(x) dx\leq P_{ABC} =0.125 a$$ $$\int_{0.5}^1 f(x) \geq P_{BCED} =0,375 a $$ 
A: 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.
