<s>[INST] If $b$ is the geometric mean of $a$ and $c$, then $b^2$ is the geometric mean of $a^2$ and $c^2$ Can someone tell me if I proved this correctly? I'm new to proofs.
\begin{align}
b^2 &= ac \\
\sqrt{b^2} &= \sqrt{ac} \\
b &= \pm \sqrt{ac} \\
b^4 &= (\sqrt{ac})^4 \\
b^4 &= a^2 c^2 \\
\sqrt[4]{b^4} &= \sqrt[4]{a^2c^2} \\
b &= \pm \sqrt{ac}
\end{align}
 [/INST] I would write the proof as follows:
If b is the geometric mean of a and c, then $b=\sqrt{ac}$.  Therefore $b^2=ac=\sqrt{a^2c^2}$, so $b^2$ is the geometric mean of $a^2$ and $c^2$.
Note that the converse is not necessarily true. If $b^2=\sqrt{a^2c^2}$, b could be negative, or a or c could be negative, in which case b would not be the geometric mean of a and c.
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