= Solution
For a faithful transitive ordered action, primitivity here means that the only invariant equivalence relations with convex classes are equality and the universal relation. The trichotomy is: a regular Archimedean action, an order two-transitive action, or a periodic action. In the regular case the chain is an ordered additive subgroup of $\mathbb R$, acting by translations. In the periodic case there is a fixed-point-free period $z$ on the <Dedekind completion>, commuting with the action and having cofinal and coinitial integer iterates; inside a period interval the relevant stabilizer acts order two-transitively. These are the cases of the <McCleary trichotomy theorem>.
\b[Only the regular Archimedean case satisfies the inequality.] It is a <linearly ordered group>. For positive $a,b$, if $a\le b$, then $ab\le b^2\le b^2a^2$; if $b\le a$, then $ab\le a^2\le b^2a^2$. Taking $a=|f|$, $b=|g|$ proves the displayed inequality.
To exclude an order two-transitive action, choose $x<y<z<t$. Choose an automorphism taking $(x,y)$ to $(y,z)$ and replace it by its pointwise <join> with the identity, giving positive $a$ with $xa=y$, $ya=z$. Similarly obtain positive $b$ with $xb=x$, $yb=t$. Then
$$
xab=t>z=xb^2a^2.
$$
Thus $ab\not\le b^2a^2$, while $|a|=a$ and $|b|=b$. The same construction inside a period interval excludes the periodic case, using the order two-transitive stabilizer supplied by the classification.
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