Question

高二数学证明计算题

已知:a>0,b>0,a+b=1
求证:　(根号下的a+2分之1)+(根号下的b+2分之1)≤2

Answer 1

a+b=1 a+1/2+b+1/2=2
[√(a+1/2)]^2+[√(b+1/2)]^2=2
由均值不等式得
2√(a+1/2)√(b+1/2)≤[√(a+1/2)]^2+[√(b+1/2)]^2=2
[√(a+1/2)+√(b+1/2)]^2=2+2√[(a+1/2)(b+1/2)]≤4
√(a+1/2)+√(b+1/2)≤2

Answer 2

a+b>=2根号下（a*b）
推导出ab<=1/4
所以ab+1/2+1/4<=1 =>ab+1/2(a+b)+1/4<=1 => (a+1/2)(b+1/2)<=1
=> 根号下的[(a+1/2)(b+1/2)]<=1 => a+b+1/2+1/2+2*根号下[(a+1/2)(b+1/2)]<=4
=> 结论

Answer 3

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Answer 4

绝对来的及，理解，总结，记忆，练习，没问题，好好干

Answer 5

令x= √(a+1/2), y= √(b+1/2),
则a =x^2 -1/2, b =y^2 -1/2.
所以x^2+y^2=4.

又因为(x+y)^2=x^2+y^2+2xy
<=x^2+y^2+(x^2+y^2)
=4,
所以 x+y<=2.
所以 ...