\(B=x^3+y^3+xy\)
\(B=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)
\(B=2x^2-2xy+2y^2+xy\)
\(B=2x^2-xy+2y^2\)
Ta có:\(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2};xy\le\dfrac{\left(x+y\right)^2}{4}\)(tự cm)
\(\Rightarrow x^2+y^2\ge2;xy\le1\)
\(\Rightarrow B\ge3\)
"="<=>x=y=1
Ta có:
\(B=x^3+y^3+xy\)
\(\Leftrightarrow B=x^3+\left(2-x\right)^3+x\left(2-y\right)\) ( Vì x + y = 2)
\(\Leftrightarrow B=x^3+8-12x+6x^2-x^3+2x-x^2\)
\(\Leftrightarrow B=5x^2-10x+8\)
\(\Leftrightarrow B=5(x^2-2x+\dfrac{8}{5})\)
\(\Leftrightarrow B=5(x^2-2x+1+\dfrac{3}{5})\)
\(\Leftrightarrow B=5(x^2-2x+1)+3\)
\(\Leftrightarrow B=5(x-1)^2+3\)
Ta thấy:
\(5(x-1)^2\ge0\forall x\)
\(\Rightarrow\)\(5(x-1)^2+3\ge3\)
hay B \(\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow x=y=1\)
Vậy Min B = 3 tại x =y= 1
