\(x^2+y^2+1\ge xy+x+y\)
\(\Leftrightarrow2\left(x^2+y^2+1\right)\ge2\left(xy+x+y\right)\)
\(\Leftrightarrow x^2-2xy+y^2+y^2-2y+1+x^2-2x+1\ge0\)\(\Leftrightarrow\left(x-y\right)^2-\left(y-1\right)^2-\left(x-1\right)^2\ge0\)
Đúng với mọi x , y
Đẳng thức xảy ra khi \(\left[{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-y=0\\y-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\y=1\\x=1\end{matrix}\right.\Rightarrow x=y=1\)
b, \(A=\dfrac{x-2}{x^3-x^2-x-2}=\dfrac{x-2}{x^3-2x^2+x^2-2x+x-2}\)
\(=\dfrac{x-2}{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}\)
\(=\dfrac{x-2}{\left(x^2+x+1\right)\left(x-2\right)}=\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)
Ta có: \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
Dấu " = " xảy ra khi \(\left(x+\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{-1}{2}\)
Vậy \(MAX_A=\dfrac{4}{3}\) khi \(x=\dfrac{-1}{2}\)
a) x2 + y2 + 1 \(\ge x.y+x+y\) \(\Leftrightarrow\) x2+y2 +1-x.y-x-y\(\ge0\)
\(\Leftrightarrow\) 2x2 + 2y2 + 2 - 2xy - 2x-2y\(\ge0\)
\(\Leftrightarrow\) ( x2 + y2 - 2 xy ) + ( x2 + 1 - 2xy ) + ( y2+1- 2y) \(\ge0\)
\(\Leftrightarrow\) ( x-y )2 + ( x-1)2 + ( y-1)2 \(\ge0\) ( Bất đẳng thức luôn đúng )
