Question

Bài 1 Cho x+y=2.Chứng minh rằng xy\(\le\)1

Answer 1

\(\text{Ta có: }x+y=2\Rightarrow x=2-y\text{ }\)

\(\Rightarrow xy=\left(2-y\right).y=2y-y^2=-y^2+2y-1+1\)

\(=-\left(y^2-2y+1\right)+1=-\left(y^2-y-y+1\right)+1\)

\(=-\left[y.\left(y-1\right)-\left(y-1\right)\right]+1=-\left(y-1\right)\left(y-1\right)+1=-\left(y-1\right)^2+1\)

\(\text{Vì }\left(y-1\right)^2\ge0\text{ nên: }-\left(y-1\right)^2\le0\Rightarrow-\left(y-1\right)^2+1\le1\)

\(\text{Vậy }xy\le1\text{ tại }y-1=0\Rightarrow y=1\Rightarrow x=2-1=1\)

Answer 2

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