Ta có: \(\left(x-y\right)^2\ge0\forall x\)
\(\left|2y-1\right|\ge0\forall y\)
\(\Rightarrow\left(x-y\right)^2+\left|2y-1\right|\ge0\forall x,y\)
\(\Rightarrow\left(x-y\right)^2+\left|2y-1\right|+3\ge3\forall x,y\)
\(\Rightarrow B\ge0\)
Dấu \("="\) xảy ra khi:
\(\left(x-y\right)^2=0;\left|2y-1\right|=0\)
Với \(\left|2y-1\right|=0\Rightarrow y=\frac{1}{2}\)
Thay \(y=\frac{1}{2}\) vào:
\(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)
Vậy \(Min_B=3\) khi \(x=\frac{1}{2}\); \(y=\frac{1}{2}.\)
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