Proof Methods

Exercise 2.b

Exercise

The geometric mean of two positive real numbers $x$ and $y$ is $\sqrt{xy}$, whereas their harmonic mean equals $\frac{2xy}{x + y}$. Prove that the geometric mean of $x$ and $y$ is at least their harmonic mean.

Proof

Note that we need to prove

$$\sqrt{xy} \ge \frac{2xy}{x + y}$$

Suppose, for the sake of contradiction, that

$$\sqrt{xy} < \frac{2xy}{x + y}$$

Divide both sides by $\sqrt{xy}$ (valid since $x,y>0$ thus $\sqrt{xy}>0$) and obtain

$$1 < \frac{2\sqrt{xy}​}{x + y}$$

Multiply both sides by $x+y$ (valid since $x,y>0$ thus $x+y>0$) and obtain

$$x+y < 2\sqrt{xy}$$

Finally, we subtract $2\sqrt{xy}$ and factor the square to obtain

$$(\sqrt{x}-\sqrt{y})^2 < 0$$

This is a contradiction since the square of any real number (in this case $\sqrt{x}-\sqrt{y}$) is non-negative