The geometric mean is always less than or equal to the arithmetic mean for any two real numbers. This can be proven using the Cauchy-Schwartz inequality, which states that, for any two vectors:
\[ \|\vec{u} + \vec{v}\| \le \|\vec{u}\| + \|\vec{v}\| \]
The arithmetic mean for any two real numbers \( a \) and \( b \) is:
\[ \frac{a+b}{2} \]
And the geometric mean for any two real numbers \( a \) and \( b \) is:
\[ \sqrt{ab} \]
We use two vectors, \( u \) and \( v \):
\[ \vec{u} = \begin{bmatrix} \sqrt{a}\\ \sqrt{b} \end{bmatrix}, \vec{v}=\begin{bmatrix} \sqrt{b}\\ \sqrt{a} \end{bmatrix} \]
Now apply the inequality:
\[ \|\vec{u} + \vec{v}\| \le \|\vec{u}\| + \|\vec{v}\| \]
\[ \| \begin{bmatrix} \sqrt{a}\\ \sqrt{b} \end{bmatrix} + \begin{bmatrix} \sqrt{b}\\ \sqrt{a} \end{bmatrix} \| \le \| \begin{bmatrix} \sqrt{a}\\ \sqrt{b} \end{bmatrix} \| + \| \begin{bmatrix} \sqrt{b}\\ \sqrt{a} \end{bmatrix}\| \]
\[ \| \begin{bmatrix} \sqrt{a} + \sqrt{b}\\ \sqrt{b} + \sqrt{a}\end{bmatrix} \| \le \| \begin{bmatrix} \sqrt{a}\\ \sqrt{b} \end{bmatrix} \| + \| \begin{bmatrix} \sqrt{b}\\ \sqrt{a} \end{bmatrix}\| \]
\[ \sqrt{(\sqrt{a} + \sqrt{b})^2 + (\sqrt{\sqrt{b}+\sqrt{a}})^2} \le \sqrt{\sqrt{a}^2 + \sqrt{b}^2} + \sqrt{\sqrt{b}^2 + \sqrt{a}^2} \]
\[ \sqrt{2 \cdot (\sqrt{a} + \sqrt{b})^2} \le 2 \cdot \sqrt{a + b} \]
\[ 2 \cdot (\sqrt{a} + \sqrt{b})^2 \le 4 \cdot (a + b) \]
\[ (\sqrt{a} + \sqrt{b})^2 \le 2 \cdot (a + b) \]
\[ a + 2\sqrt{a}\sqrt{b} + b \le 2a + 2b \]
\[ 2\sqrt{ab} \le a + b \]
\[ \sqrt{ab} \le \frac{a + b}{2} \]
The last equation demonstrates that the geometric mean is always less than or equal to the arithmetic mean.