This came from Paul Zeitz's book called The Art and Craft of Problem Solving. It's problem 5.5.40 on p.203. he

Question

Let a,b,c,d >= 0. Prove that

[3( a + b + c)]^ (1/2) >= (a) ^ (1/2) + (b) ^ (1/2) + (c) ^ (1/2)

Answer 1

First, for r >= 0 and s >= 0,

(r + s)^2 = r^2 + s^2 + 2rs

               = (r^2 + s^2 - 2rs) + 4rs

               = (r - s)^2 + 4rs

Since (r - s)^2 >= 0, then (r + s)^2 >= 4rs, and since both sides of this inequality are positive, we can take the square root of both sides and get

r + s >= 2(rs)^(1/2)

3(a + b + c) = a + b + c + (a + b) + (a + c) + (b + c)

         >= a + b + c + 2(ab)^(1/2) + 2(ac)^(1/2) + 2(bc)^(1/2)

           = ((a)^(1/2) + (b)^(1/2) + (c)^(1/2))^2

Again, since both sides of this inequality are positive we can take the square root of both sides and get

(3(a + b+ c))^(1/2) >= (a)^(1/2) + (b)^(1/2) + (c)^(1/2)