Problem 7

Prove that if $0 < a < b$ then \[ \begin{align*} a < \sqrt{ab} < \frac{a + b}{2} < b \\ \end{align*} \] Notice that the inquality $\sqrt{ab} \leq \frac{a + b}{2}$ holds for all $a, b \geq 0$. So, first we have to prove that $a < \sqrt{ab}$. \[ \begin{align*} a &< b \\ aa &< ab \\ a &< \sqrt{ab} \\ \end{align*} \] Next, we prove $\frac{a + b}{2} < b$. \[ \begin{align*} a &< b \\ a + b &< 2b \\ \frac{a + b}{2} &< b \\ \end{align*} \] With the given inquality we have the proof that this statement is true.

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