- Given that \(a > 0\), \(b > 0\), prove that \[\begin{equation*} \frac{a}{b} + \frac{b}{a} \ge 2. \end{equation*}\]

As the final inequality is always true, the first inequality must also be true.

- Given that \(x > 0\), \(y > 0\), \(z > 0\) and that \(x + y + z = 3\), prove that \[\begin{equation*} \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \ge 3. \end{equation*}\]

There are three common means for \(n\) positive numbers, \(x_1 + \dotsb + x_n\). We have;

the Arithmetic Mean = \(\dfrac{x_1 + \dotsb + x_n}{n}\), the Geometric Mean = \(\sqrt[n]{x_1\dotsb x_n}\), and the Harmonic Mean, \(\dfrac{n}{\dfrac{1}{x_1} + \dotsb + \dfrac{1}{x_n}}.\)

Now it’s always true that \(AM \geq GM \geq HM\). The first part of the question here is the \(AM \geq GM\) inequality for the numbers \(\dfrac{a}{b}\) and \(\dfrac{b}{a}\).

Since the harmonic mean is less than or equal to the arithmetic mean, we have \[\begin{equation*} \frac{n}{\frac{1}{x_1} + \dotsb + \dfrac{1}{x_n}} \le \dfrac{1}{n} \left( x_1 + \dotsb + x_n \right). \end{equation*}\] We can rearrange this to become \[\begin{equation*} \frac{n^2}{x_1 + \dotsb + x_n} \le \frac{1}{x_1} + \dotsb + \frac{1}{x_n}. \end{equation*}\]So the result we proved in the second part of the problem was a special case of this inequality.