The following system of ODEs arises when studying Ricci flow on 3-manifolds: $$ \frac{dm_1}{dt} = m_1^2+m_2m_3 \\ \frac{dm_2}{dt} = m_2^2+m_1m_3 \\ \frac{dm_3}{dt} = m_3^2+m_1m_2 \\ $$
Going back over Hamilton's 1986 paper I realised I didn't understand the first step of his reasoning:
Note that $$\frac{d}{dt}(m_2 - m_1) = (m_2-m_1)(m_2+m_1-m_3)$$ so that if $m_1 \le m_2$ to start it remains so.
The only extra context required is that $m_2 \le m_3$ at the initial time. How can we make this conclusion? It's not as simple as $m_2 - m_1$ being non-decreasing; choosing a large value for $m_3$ makes this clear. I initially thought to rewrite as $$\frac{d}{dt}\log(m_2-m_1) = m_2 + m_1 - m_3$$ but at this stage of the analysis I see no reason why $\int (m_2 + m_1 - m_3) dt$ should not fly off to $-\infty$; indeed we have blowup in finite time for the "similar" equation $\frac{df}{dt}=f^2$.
I have a feeling that I'm missing something very easy and will shortly be quite embarrassed, but it's been niggling at me.
And after thinking for a little while I realise that the formula I noted earlier:$$\log(m_2(t) - m_1(t)) = \log(m_2(0)-m_1(0)) + \int_0^t (m_2 + m_1 - m_3)$$ is essentially all I need.
If $T<\infty$ is the first time such that $m_2(T) = m_1(T)$, then $\log(m_2-m_1)$ is well defined on $[0,T)$ and converges to $-\infty$ as $t\nearrow T$. This implies $\liminf_{t\nearrow T} (m_2 + m_1 - m_3) = -\infty$ and thus at least one of the $m_i$ has a discontinuity at $T$. Thus $m_2 \ge m_1$ so long as the solution exists.
That is, a blowup in finite time can occur, but I don't actually care! Sanity checks welcome.