已知a,b∈R,求證:a2+b2≥ab+a+b-1.
已知a,b∈R,求證:a2+b2≥ab+a+b-1.
數(shù)學人氣:385 ℃時間:2020-07-25 20:42:23
優(yōu)質(zhì)解答
證明:(a
2+b
2)-(ab+a+b-1)
=
(2a
2+2b
2-2ab-2a-2b+2)
=
[(a
2-2ab+b
2)+(a
2-2a+1)+(b
2-2b+1)]
=
[(a-b)
2+(a-1)
2+(b-1)
2]≥0,
∴a
2+b
2≥ab+a+b-1.
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