How Can You Prove the Inequality Sqrt(ab) <= (a+b)/2?

[lovemake1]

Homework Statement

Given:0 <= a <= b
a <= Sqrt(ab) <= (a+b)/2 <= b

Homework Equations

The Attempt at a Solution

The only problem I am having prooving this inequality is Sqrt(ab) <= (a+b)/2.
I have an idea but I am not sure if it validates.
can i do this.. ? (a+b)/2 - sqrt(ab) >= 0
if it is greater than 0 [ i get an answer of 1/4 (a-b)^2 >= 0 ] , (a+b)/2 must be greater than sqrt(ab). given that
both sqrt(ab) and (a+b)/2 are >= 0 since 0 < a < b.

Is my reasoning correct? or wrong
please help !

 

[nate808]

Your reasoning is correct. To prove that Sqrt(ab) <= (a+b)/2, you can use the fact that the square root function is increasing. This means that if x < y, then Sqrt(x) < Sqrt(y). Since a < b, we know that Sqrt(a) < Sqrt(b). We can then multiply both sides by Sqrt(b) to get Sqrt(ab) < b. Similarly, we can multiply both sides by Sqrt(a) to get Sqrt(ab) < a. Therefore, Sqrt(ab) is between a and b, and since (a+b)/2 is the average of a and b, it must be greater than or equal to Sqrt(ab).

Another way to prove this inequality is to use the AM-GM inequality, which states that for any two positive numbers a and b, their arithmetic mean (a+b)/2 is greater than or equal to their geometric mean Sqrt(ab). Since a and b are both positive (since 0 < a < b), this inequality holds and we can conclude that (a+b)/2 >= Sqrt(ab).

Overall, your reasoning is correct and you can use either method to prove the inequality.