Question

Cho:x,y thuộc R.CMR:\(x^2+y^2+1\ge xy+x+y\)

Answer 1

\(x^2+y^2+1\ge xy+x+y\)

<=>\(2\left(x^2+y^2+1\right)\ge2\left(xy+x+y\right)\)

<=>\(2x^2+2y^2+2\ge2xy+2x+2y\)

<=>\(2x^2+2y^2+2-2xy-2x-2y\ge0\)

<=>\(\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2-2y+1\right)\ge0\)

<=>\(\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2\ge0\) (luôn đúng)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y=0\\x-1=0\\y-1=0\end{cases}\Leftrightarrow x=y=1}\)

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