My course problem booklet (Mathematics BSc, second year module in analysis, unpublished) has:
Let $a \geq b \geq 0$. We define sequences $(a_n)$ and $(b_n)$ by taking $a_1$ and $b_1$ to be $a$ and $b$ respectively, and requiring that for $n \geq 1$, $$ a_{n+1} = \frac{1}{2}(a_n+b_n) \hspace{1em} \text{and} \hspace{1em} b_{n+1} = \sqrt{a_nb_n}. $$ In other words, $a_{n+1}$ is the arithmetic mean of $a_n$ and $b_n$ while $b_{n+1}$ is their geometric mean.
Prove that $b_n \leq b_{n+1} \leq a_{n+1} \leq a_n$.
For the left inequality, the solution booklet has:
$$ \begin{aligned} b_{n+1}^2 - b_n^2 & = a_nb_n - a_{n-1}b_{n-1} \\ & = \frac{1}{2}(a_{n-1}+b_{n-1})\sqrt{a_{n-1}b_{n-1}} - a_{n-1}b_{n-1} \\ & = \sqrt{a_{n-1}b_{n-1}}\left(\frac{1}{2}(a_{n-1}+b_{n-1})-\sqrt{a_{n-1}b_{n-1}}\right) \geq 0 \end{aligned} $$ by the theorem of the means. So $b_{n+1}^2 \geq b_n^2$ and taking square roots we have $b_{n+1} \geq b_n$ for all $n$.
Now, I see that the Theorem of the Means (ToM) gives us that $$ \begin{aligned} & \; \frac{1}{2}(a_{n-1}+b_{n-1}) \geq \sqrt{a_{n-1}b_{n-1}} \\ \therefore & \; \frac{1}{2}(a_{n-1}+b_{n-1})-\sqrt{a_{n-1}b_{n-1}} \geq 0 \end{aligned} $$ but we want $$ \sqrt{a_{n-1}b_{n-1}}\left(\frac{1}{2}(a_{n-1}+b_{n-1})-\sqrt{a_{n-1}b_{n-1}}\right) \geq 0 $$ and that also requires $$ \sqrt{a_{n-1}b_{n-1}} \geq 0. $$ Unless we reason $$ \begin{aligned} & \; \text{this is real analysis} \\ \therefore & \; \sqrt{a_{n-1}b_{n-1}} \in \mathbb{R} \\ \therefore & \; a_{n-1}b_{n-1} \geq 0 \\ \therefore & \; \sqrt{a_{n-1}b_{n-1}} \geq 0, \end{aligned} $$ this would seem to require $$ a_{n-1},b_{n-1} \leq 0 \hspace{1em} \text{or} \hspace{1em} a_{n-1},b_{n-1} \geq 0 $$ and to get the expected $a_{n-1},b_{n-1} \geq 0$ from the given $a \geq b \geq 0$ would seem to require us to assume the inequalities we're trying to prove.
Thoughts?