Đặt \(A=x^2+y^2-x-y-xy\)
\(\Leftrightarrow2A=2x^2-2y^2-2xy-2x-2y\)
\(\Leftrightarrow2A=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2-2y+1\right)-2\)
\(\Leftrightarrow2A=\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2-2\)
Vì \(\left(x-y\right)^2\ge0\forall\) x,y;\(\left(x-1\right)^2\ge0\forall x\);\(\left(y-1\right)^2\ge0\forall\)y nên \(2A\ge-2\)
Hay \(A\ge-1\)
Dấu bằng xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x-y=0\\x-1=0\\y-1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=y\\x=1\\y=1\end{cases}}\)\(\Leftrightarrow x=y=1\)
Vậy Min A=-1 khi x=y=1
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