How does one prove or disprove this integral inequality for a $C^1([0,1])$ function with zero average?

Question

By zero average, I mean $\int_0^1 f(x) dx = 0$. The inequality is $$ 2 \int_0^1 [f(x)]^2 dx \le \left(\int_0^1 |f(x)|dx\right)\left(\int_0^1 |f'(x)|dx\right). $$ Cauchy-Schwarz hasn't led me anywhere, and Poincare's Inequality doesn't apply in an obvious fashion.

Answer 1

Note that $f'$ (and thus $|f'|$) is Riemann-integrable on $[0,1].$ Then, using the condition $$\int_0^1 f(x) \,dx = 0$$ we see $$\int_0^1 f(x)^2\, dx = \int_0^1 f(x)(f(x) - f(0))\, dx = \int_0^1 f(x)\int_0^xf'(t)\,dt\, dx \le \int_0^1 |f(x)| \int_0^x|f'(t)|\,dt\, dx.$$Similarly, $$\int_0^1 f(x)^2\, dx = \int_0^1 f(x)(f(x) - f(1))\, dx \le \int_0^1 |f(x)|\, \int_x^1|f'(t)|\,dt\, dx.$$ Adding the two inequalities gives the desired result.