I'm taking a proofs class and the textbook says to do this problem:
For all real $ a, b > 0 $, show
$ \dfrac{2ab}{a + b} \leq \sqrt{ab} \leq \dfrac{a + b}{2} \leq \sqrt{\dfrac{a^2 + b^2}{2}} $
It recommends solving each inequality separately and putting them together.
Rather frustratingly, I was able to do the first two inequalities but not the last. Here's how:
For reals r, s, the following equation holds in general:
$$ 0 \leq (r-s)^2 $$ $$ 0 \leq r^2 - 2rs + s^2 $$ $$ 2rs \leq r^2 + s^2 $$ $$ 2r^2s^2 \leq rs(r^2 + s^2) $$
Let $ a = r^2 $ and $ b = s^2. $
$$ 2ab \leq \sqrt{ab}(a + b) $$ $$ \dfrac{2ab}{a + b} \leq \sqrt{ab} $$
The other is:
$$ 0 \leq (r - s)^2 $$ $$ 0 \leq \dfrac{r^2 - 2rs + s^2}{2} $$ $$ rs \leq \dfrac{r^2 + s^2}{2} $$
Using the same substitution,
$$ \sqrt{ab} \leq \dfrac{a + b}{2} $$
I will find out the answer in class later, but it bothers me. As for the last one, how is it done?
Not too hard actually! Once I realized everything was by definition positive...
$$ 0 \leq (a - b)^2 $$ $$ \dfrac{-(a - b)^2}{2} \leq 0 $$ $$ \dfrac{-a^2 + 2ab - b^2}{4} \leq 0 $$ $$ \dfrac{a^2 + 2ab + b^2}{4} - \dfrac{a^2 + b^2}{2} \leq 0 $$ $$ 0 \leq (\dfrac{a + b}{2})^2 \leq \dfrac{a^2 + b^2}{2} $$
Since both sides are positive, we can take a square root by the square root lemma (it's proved in my book):
$$ \dfrac{a + b}{2} \leq \sqrt{\dfrac{a^2 + b^2}{2}} $$
By the way, this is the harmonic mean of a and b, the geometric mean of a and b, the arithmetic mean of a and b, and the root mean square of a and b.