a) Vì \(\left|x-\frac{1}{3}\right|\ge0\forall x\)\(\Rightarrow\left|x-\frac{1}{3}\right|+\frac{1}{4}\ge\frac{1}{4}\forall x\)
hay \(P\ge\frac{1}{4}\)
mà \(\frac{1}{4}>\frac{1}{5}\)\(\Rightarrow P>\frac{1}{5}\)
b) Ta có: \(P\ge\frac{1}{4}\)( chứng minh phần a )
Dấu " = " xảy ra \(\Leftrightarrow x-\frac{1}{3}=0\)\(\Leftrightarrow x=\frac{1}{3}\)
Vậy \(minP=\frac{1}{4}\Leftrightarrow x=\frac{1}{3}\)