University Analysis 2 Integral Inequality

We can prove this by only assuming that $f(a)=0$. By Holder's inequality we have 
$$\int_a^b|f(x)f'(x)|\mathrm{d}x \le \left( \int_{a}^{b}|f(x)|^2 \mathrm{d}x \right)^{\frac{1}{2}}\left( \int_{a}^{b}|f'(x)|^2 \mathrm{d}x \right)^{\frac{1}{2}}\tag{1}$$
Since $f(a)=0$, $$f(x)=\int_{a}^{x}f'(t)\mathrm{d}t \tag{2}$$
From (2) we can write
$$|f(x)| \le \int_a^x|f'(t)|\mathrm{d}t \le (x-a)^{\frac{1}{2}}\left( \int_a^x |f'(t)|^2 \mathrm{d}t \right)^{1/2}$$
Thus 
$$\int_a^b|f(x)|^2\mathrm{d}x \le \int_a^b(x-a)\int_a^x|f'(t)|^2\mathrm{d}t\mathrm{d}x \le \int_a^b |f'(t)|^2\mathrm{d}t\int_a^b(x-a)\mathrm{d}x $$
\[=\int_a^b |f'(t)|^2\mathrm{d}t \cdot (\frac{b-a}{2})^2\]
Hence 
$$(\int_a^b|f(x)|^2\mathrm{d}x)^{\frac{1}{2}} \le  (\int_a^b |f'(t)|^2\mathrm{d}t)^{\frac{1}{2}}(\frac{b-a}{2})  \tag{3}.$$

Substituting (3) in (1) we get 

$$\int_a^b|f(x)f'(x)|\mathrm{d}x \le \frac{b-a}{2}  \int_{a}^{b}|f'(x)|^2 \mathrm{d}x.$$