Question

已知a>0b>0 且a+b=1 求证 根号(a+1/2)+根号(b+1/2)≤2

Answer 1

由公式(x+y)^2≤2(x^2+y^2),得x+y≤(2(x^2+y^2))^(1/2)这里,x=(a+1/2)^(1/2),y=(b+1/2)^(1/2),即(a+1/2)^(1/2)+(b+1/2)^(1/2)≤(2(a+1/2+b+1/2))^(1/2)=(2a+1+2b+1)^(1/2)=(2(a+b)+2)^(1/2)=4^(1/2)=2所以,(a+1/2)^(...