问题
填空题
已知a>0,b>0,a+b=1,则
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答案
∵a>0,b>0,a+b=1,∴
+1 a
=(a+b)(1 b
+1 a
)=2+1 b
+b a
≥2+2a b
=4,当且仅当a=b=
×b a a b
时取等号.1 2
∴
+1 a
的取值范围是[4,+∞).1 b
故答案为[4,+∞).
已知a>0,b>0,a+b=1,则
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∵a>0,b>0,a+b=1,∴
+1 a
=(a+b)(1 b
+1 a
)=2+1 b
+b a
≥2+2a b
=4,当且仅当a=b=
×b a a b
时取等号.1 2
∴
+1 a
的取值范围是[4,+∞).1 b
故答案为[4,+∞).
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