求不等式1/(ka+b)+1/(kb+a)>=A/(a+b)在0
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![求不等式1/(ka+b)+1/(kb+a)>=A/(a+b)在0](/uploads/image/z/6589665-9-5.jpg?t=%E6%B1%82%E4%B8%8D%E7%AD%89%E5%BC%8F1%2F%28ka%2Bb%29%2B1%2F%28kb%2Ba%29%3E%3DA%2F%28a%2Bb%29%E5%9C%A80)
求不等式1/(ka+b)+1/(kb+a)>=A/(a+b)在0
求不等式1/(ka+b)+1/(kb+a)>=A/(a+b)在0
求不等式1/(ka+b)+1/(kb+a)>=A/(a+b)在0
依柯西不等式得
1/(ka+b)+1/(kb+a)
≥(1+1)^2/(ka+kb+a+b)
=4/(k+1)(a+b).
要使原不等式恒成立,则
A/(a+b)≤4/(k+1)(a+b)
→A=4/(k+1).
考虑到0≤k≤2011,知
当k=0时,
所求A的最大值为:4.