Question

若正数a,b满足a+b=1,求证:根号下2a+1加上根号下2b+1小于等于2根号2?

Answer 1

（√（2 a+1）+√(2b+1)）的平方
=2a+2b+1+2√(2a+1)×√(2b+1) ∵ a+b=1∴b=1-a
=4+2√(2a+1)(3-2a)
∵ (2a+1)(3-2a)≥0∴-1/2≤a≤3/2 a>0 b>0,a+b=1 则 0≤a<1
∴配方 画数轴得3< (2a+1)(3-2a) ≤4
∴4+2√3<（√（2 a+1）+√(2b+1)）的平方≤4+2√4=8
∴根号下2a+1加上根号下2b+1小于等于2根号2

Answer 2

要证明:√2a+1 +√2b+1 <= 2√2
即要证明: (√2a+1 +√2b+1)2 <= 8
2a+1 + 2b+1 +√4ab+2a+2b <= 8
因为a+b=1
所以 2+2+2√4ab+2a+2b+1 <= 8
√4ab+2a+2b+1<= 2
两边平方根 :4ab+3 <=4
4ab <= 1
4ab <= a2+2ab+b2
0 <= a2-2ab+b2
0<=(a-b)2
所以得证