this question pisses me off (1 Viewer)

[blackbunny]

If a and b are any two non-negative real numbers prove that:


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

i have no idea att all wtf?

 

[BlackJack]

To understand it, try squaring both sides... think about the relation btwn (a+b)^2 and ab... then write the proof properly.

 

[redruM]

(a+b)/2 >= #(ab)
(a+b)²/2² >= {#(ab)}²
(a+b)²/4 >= ab
(a+b)² >= 4ab
a² + 2ab + b² >= 4ab
a² - 2ab + b² >= 0
(a - b)² >= 0

difference of any 2 non-negative numbers squared will always be >= 0.

 

[~ ReNcH ~]

(sqrt a - sqrt b)² >= 0, a>=0 and b>=0
.'. a - 2sqrt(ab) + b >= 0
.'. a + b >= 2sqrt(ab)
.'. (a + b)/2 >= sqrt(ab)