SOLUTION TO PROBLEM 3:

Since a and b are greater than zero, we have:

When a=b:

$$\frac{a+a}{2}=\frac{2a}{2}=a;\qquad \sqrt{a\cdot a}=a;\qquad \frac{2aa}{a+a}=\frac{2a^2}{2a}=a.$$

When $a\neq b$:
We know that $(\sqrt{a}-\sqrt{b})^2>0$, and thus $a-2\sqrt{ab}+b>0$. Hence

$$a+b>2\sqrt{ab}\quad\mbox{ and }\quad \frac{a+b}{2}>\sqrt{ab}.$$

This establishes the first inequality. Next note that

$$\sqrt{ab}\cdot\frac{a+b}{2}>\sqrt{ab}\cdot\sqrt{ab}=ab,$$

and thus

$$\sqrt{ab}\cdot (a+b)>2ab,$$

so

$$\sqrt{ab}>\frac{2ab}{a+b}.$$

This establishes the second inequality.

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Mon Sep 11 09:30:12 CDT 2000