Từ \(3x+y-1=0\Rightarrow y=1-3x\)
\(\Rightarrow P=3x^2+y^2=3x^2+\left(1-3x\right)^2=3x^2+9x^2-6x+1\)
\(=12x^2-6x+1=12\left(x-\dfrac{1}{4}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\) có GTNN là \(\dfrac{1}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{4}\)