由于|a+b|在a、b同号时值最大,故可设a>0,b>0(a、b小于0时结果一样);
(a*a+1)(b*b+1) =a²b²+a²+b²+1=a²b²+(a²-a+1/4)+(b²-b+1/4)+a+b+1/2
=a²b²+(a-1/2)²+(b-1/2)²+a+b+1/2>a+b=|a+b|
若a<0,b<0,则为
(a*a+1)(b*b+1) =a²b²+a²+b²+1=a²b²+(a²+a+1/4)+(b²+b+1/4)-a-b+1/2>-a-b=|a+b|
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