求证：a²+b²≥ab+a+b-1
a²+b²-ab+-a-b+1怎么又等于1/2(a²-2ab+b²)+1/2(a²-2a+1)+1/2(b²-...
a²+b²-ab+-a-b+1
怎么又等于1/2(a²-2ab+b²)+1/2(a²-2a+1)+1/2(b²-2b+1) 呢？ 展开
怎么又等于1/2(a²-2ab+b²)+1/2(a²-2a+1)+1/2(b²-2b+1) 呢？ 展开
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a²+b²-ab-a-b+1
=(2a²+2b²-2ab-2a-2b+2)/2
=(a²-2ab+b²+a²-2a+1+b²-2b+1)/2
=[(a²-2ab+b²)+(a²-2a+1)+(b²-2b+1)]/2
=[(a-b)²+(a-1)²+(b+1)²]/2≥0
a²+b²-ab-a-b+1≥0
a²+b²≥0ab+a+b-1
=(2a²+2b²-2ab-2a-2b+2)/2
=(a²-2ab+b²+a²-2a+1+b²-2b+1)/2
=[(a²-2ab+b²)+(a²-2a+1)+(b²-2b+1)]/2
=[(a-b)²+(a-1)²+(b+1)²]/2≥0
a²+b²-ab-a-b+1≥0
a²+b²≥0ab+a+b-1
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