若正数ab满足a+b=1求证根号2a+1+根号2b+1

Question

若正数ab满足a+b=1求证根号2a+1+根号2b+1

Answer 1

√(2a+1)+√(2b+1)(记得根号下的内容要打括号括起来)
∵√(2a+1)+√(2b+1)>0
[√(2a+1)+√(2b+1)]^2
=2a+1+2b+1+2√[(2a+1)(2b+1)]
=2(a+b)+2+2√[2(a+b)+1+4ab]
=4+2√(3+4ab)
≤4+2√(3+2ab+a^2+b^2)
=4+2√[3+(a+b)^2]
=8
∴√(2a+1)+√(2b+1)≤√8=2√2
故其最大值为2√2,而2√2≈2.828