Question

Given \(a \ge 0\) and \(b \ge 0\), let \({\rm{u}} = \left( \begin{array}{l}\sqrt a \\\sqrt b \end{array} \right)\)and\({\rm{v}} = \left( \begin{array}{l}\sqrt b \\\sqrt a \end{array} \right)\). Use the Cauchy–Schwarz inequality to compare the geometric mean \(\sqrt {ab} \) with the arithmetic mean \(\frac{{\left( {a + b} \right)}}{2}\).

Short answer

The required inequality is \(2\sqrt {ab} \le a + b\).

Step-by-step solution

Use the given information

It is given that \({\rm{u}} = \left( \begin{array}{l}\sqrt a \\\sqrt b \end{array} \right)\) and \({\rm{v}} = \left( \begin{array}{l}\sqrt b \\\sqrt a \end{array} \right)\). So, according to inner product axioms, \({\left\| {\rm{u}} \right\|^2} = a + b\) and \({\left\| {\rm{v}} \right\|^2} = a + b\)and\(\left| {\left\langle {{\rm{u,v}}} \right\rangle } \right| = 2\sqrt {ab} \).

Use Cauchy Schwarz inequality

According to Cauchy Schwarz inequality, an inner product on a vector space \(V\) is a function that, to each pair of vectors \({\bf{u}}\) and \({\rm{v}}\) in \(V\), associates a real number \(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle \)and satisfies the axiom, \(\left| {\left\langle {{\bf{u}},{\rm{v}}} \right\rangle } \right| \le \left\| {\rm{u}} \right\|\left\| {\rm{v}} \right\|\) for all \({\bf{u}}\) and \({\rm{v}}\)in \(V\).

Plug the above obtained values in Cauchy Schwarz inequality, as follows:

\(\begin{array}{c}\left\langle {{\rm{u,v}}} \right\rangle \le \left\| {\rm{u}} \right\|\left\| {\rm{v}} \right\|\\2\sqrt {ab} \le \sqrt {a + b} \sqrt {a + b} \\2\sqrt {ab} \le a + b\end{array}\)

Thus, the required inequality is\(2\sqrt {ab} \le a + b\).