Answer to Question #3401 in Real Analysis for junel
If 0 < a < b, show that
1)& a < (ab)[sup]1/2[/sup] < b;
2) 1/b < 1/a.
0 < a < b Let's multiply the given inequality: a < b | x a thus a2 < ab Since a, b are positive, √(a2) < √ (ab) a < (ab)1/2 In similar way ( * b) we can obtain (ab)1/2 < b, thus a < (ab)1/2 < b. There are positive numbers a and b, such that a < b. Let's divide this inequality by a: a/a < b/a. 1 < b/a. Then divide it by b and get: 1/b < 1/a. The statement is proved.