Suppose that $f\in C^1([0,\infty),\mathbb{R})$,and $F(x)=\max_{x\leq y\leq 2x}|f(y)|$,then show that $ \int_{0}^{\infty}F(x)dx\leq \int_{0}^{\infty}|f(x)|dx+\int_{0}^{\infty}x|f'(x)|dx $
EDIT
Thanks to richard's counterexample,I realised now that the term $\dfrac{1}{x}$ should be replaced by $x$.
One try is to write $ f(x)=\frac{1}{x}\int_{x}^{2x}(f(x)-f(y))dy+\frac{1}{x}\int_{x}^{2x}f(y)dy\\ =\frac{1}{x}\int_{x}^{2x}f(y)dy-\frac{1}{x}\int_{x}^{2x}\int_{x}^{y}f'(z)dz $ then,we have the following $ \dfrac{1}{x}\int_{x}^{2x}f(y)dy=2f(2x)-f(x)+\frac{1}{x}\int_{x}^{2x}yf'(y)dy $and fact that $\left \vert \dfrac{1}{x} \displaystyle \int_{x}^{2x}f(y)dy \right \vert\leq F(x) $