Question #3401

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

Expert's answer

0 < a < b

Let's multiply the given inequality:

a < b | x a

thus

a^{2} < ab

Since a, b are positive, √(a^{2}) < √ (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.

Let's multiply the given inequality:

a < b | x a

thus

a

Since a, b are positive, √(a

a < (ab)

In similar way ( * b) we can obtain (ab)

a < (ab)

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.

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