Question

已知a>0,b>0且a+b=1.求证根号下a+1/2 +根号下b+1/2≤2

已知a>0,b>0且a+b=1.求证根号下a+1/2 +根号下b+1/2≤2

Answer 1

√(a+1/2) +√(b+1/2)≤2
两边同时平方
a+b+1+2√[(a+1/2)(b+1/2)]<=4
[(a+1/2)(b+1/2)]<=1
ab+1/4+(1/2)(a+b)<=1
ab<=(1/4)(a+b)^2<=1/4
得证

Answer 2

根号下a+1/2 +根号下b+1/2
<=[(根号a+1/2)^2+（根号b+1/2)^2]/2
=(a+1/2+b+1/2)/2
=4/2
=2（错了）。。。等一下

原题得证