Any hints to prove that (a+b)/2 >= sqrt(ab)

for all positive numbers a and b. im not asking for the entire answer, some hints on how to start with it would be great. thank you.

Start with (a-b)^2>0

Terminology: iff is short for "if and only if." I.e., one statement is equivalent to the other.

Start by squaring both sides. Note that, since a and b are positive, both sides of the inequality are also positive. From there, we use the fact that if x and y are positive, then x≥y iff x^2 ≥ y^2 (this in turn can be proved by various methods).

From there, we may proceed:
(a+b)/2 >= sqrt(ab) ...iff
(a+b) >= 2*sqrt(ab) ...iff
(a+b)^2 >= 4ab ...iff
a^2 + 2ab + b^2 ≥ 4ab ...iff
a^2 - 2ab + b^2 ≥ 0... iff
(a - b)^2 ≥ 0

Since the square of any real number is greater than or equal to zero, it follows that all the equivalent statements are true. Thus, (a+b)/2 >= sqrt(ab)

Proof of the earlier statement: for positive x,y x≥y iff x^2 ≥ y^2

x^2 ≥ y^2 iff
x^2 - y^2 ≥ 0 iff
(x+y)(x-y) ≥ 0 iff ...note that since (x+y)>0, we may divide both sides by it to get that
x-y≥0 iff
x ≥ y

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

(a+b)^2>= 4ab
(a+b)>= 2 sqrt(ab)
(a+b)/2 >= sqrt(ab)