Answer
The prove is shown below.
Work Step by Step
Let $a, b>0$. Now, the geometric mean of these numbers is $\sqrt{ab}$ while the arithmetic mean is $\dfrac{a+b}{2}$.
Note that for any real number $x$, we have $x^2\geq 0$.
Thus,
$(\sqrt{a}-\sqrt{b})^2\geq 0$
$\Rightarrow (\sqrt{a})^2+(\sqrt{b})^2-2\sqrt{a}\sqrt{b}\geq 0$.
$\Rightarrow a+b-2\sqrt{ab}\geq 0$
$\Rightarrow a+b\geq 2\sqrt{ab}$
$\Rightarrow \dfrac{a+b}{2}\geq \sqrt{ab}$
$\Rightarrow \text{Arithmetic Mean of }a\text{ and } b \geq \text{Geometric Mean of } a \text{ and }b$.
Therefore, the geometric mean $\sqrt{ab}$ is not larger than the arithmetic mean $\dfrac{a+b}{2}$ always.
