Prove that: (a+b)/2 = sqrt( a*b) When a>=0 and b>=0 ?

1 Answer
Oct 1, 2017

(a+b)/2 color(red)(>=) sqrt(ab)" " as shown below

Explanation:

Note that:

(a-b)^2 >= 0" " for any real values of a, b.

Multiplying out, this becomes:

a^2-2ab+b^2 >= 0

Add 4ab to both sides to get:

a^2+2ab+b^2 >= 4ab

Factor the left hand side to get:

(a+b)^2 >= 4ab

Since a, b >= 0 we can take the principal square root of both sides to find:

a+b >= 2sqrt(ab)

Divide both sides by 2 to get:

(a+b)/2 >= sqrt(ab)

Note that if a != b then (a+b)/2 > sqrt(ab), since then we have (a-b)^2 > 0.