<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] Some words would be nice. You seem to be counting on the agreement between the third line and the last line, but you don't point that out.  Also, the geometric mean is defined as the positive square root, so you don't need the $\pm$ signs.  I would write it
We are given $b=\sqrt{ac}$  
The geometric mean of $a^2$ and $c^2$ is $\sqrt{a^2c^2}=(\sqrt{ac})^2=b^2$
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