$\left| f\left( 0 \right) \right|=\left| b \right|,\,\,\left| f\left( 1 \right) \right|=\left| 1+a+b \right|,\,\,\left| f\left( -1 \right) \right|=\left| 1-a+b \right|$
Xét $$\left| b \right| < \frac{1}{2}\Rightarrow 0\le \left| b \right|\frac{-1}{2}\Rightarrow \left\{ \begin{align}
& 1-a+b>\frac{1}{2}-a \\
& 1+a+b>\frac{1}{2}+a \\
\end{align} \right.$$
Nếu $a\ge 0\Rightarrow \left| 1+a+b \right|>\left| \frac{1}{2}+a \right|\ge \frac{1}{2}$ hay $f\left( 1 \right)>\frac{1}{2}$
Nếu $a\le 0\Rightarrow \left| 1-a+b \right|>\left| \frac{1}{2}-a \right|\ge \frac{1}{2}$ hay $f\left( -1 \right)>\frac{1}{2}$
Xét $0\le b < \frac{1}{2}\Rightarrow \left\{ \begin{align}
& 1-a+b\ge 1-a \\
& 1+a+b\ge 1+a \\
\end{align} \right.$
Nếu $a\ge 0\Rightarrow \left| 1+a+b \right|\ge \left| 1+a \right|\ge 1$ hay $f\left( 1 \right)>\frac{1}{2}$
Nếu $a\le 0\Rightarrow \left| 1-a+b \right|\ge \left| 1-a \right|\ge 1$ hay $f\left( -1 \right)>\frac{1}{2}$
Vậy ta có điều phải chứng minh.