let a,b are positiv real such abc=1
prove that a+b+c>=sqrt(a)+sqrt(b)+ sqrt(c)
Is there a digit 2 in the problem?
If not, the inequality does not hold for some positive a,b, a+b=1.
If there is a digit 2, then applying the inequlity mentioned above, substitute t with a and b to get two inequality, add them up, the result follows immediately.