Question

If 0 < a < b, show that
1)& a < (ab)[sup]1/2[/sup] < b;
2) 1/b < 1/a.

Answer 1

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.