咳咳!答案来了!
左边化为:1/2*(a+b)*[1/2*(a+b)+1] 右边化为:√ab(√a+√b)
利用均值不等式得:1/2*(a+b)≥√ab
所以只需证明:1/2*(a+b)+1≥√a+√b
1/2*(a+b)+1- (√a+√b)
=1/2(a+1)-√a+ 1/2(b+1)-√b
∵ √a =√a*1<=1/2(a+1)
√b =√b*1<=1/2(b+1)
∴1/2(a+1)-√a+ 1/2(b+1)-√b≥0
∴1/2*(a+b)+1≥√a+√b
∴a+b)^2/4+(a+b)/2≥a√b +b√a