Squared bound from comparison at values ($Vpq\ge Vx\text{ for all values of }x⟹p^2{q}^2\ge x$)

Let $p,q,x\ge 1$ in a lattice-ordered group. Under the displayed hypothesis, suppose $d=(x{q}^{-2}{p}^{-2})\vee 1>1$ and choose a value $P$ of $d$. If $x\in P$, then $1\le d\le x$ contradicts its omission. Otherwise extend $P$ to a value $V$ of $x$. The right-coset test is $Vu\le Vv⟺(u{v}^{-1})\vee 1\in V$. If $p\notin V$, put $c=(x{q}^{-2}{p}^{-1})\vee 1\in V$. Then $(p{c}^{-1})\vee 1\notin V$, while its inverse positive part is $(c{p}^{-1})\vee 1=d$. These parts have meet one, so primeness of $P$ forces $d\in P$. If $p\in V$, put $c=(x{q}^{-1})\vee 1\in V$ and $t=p^{2}q\notin V$; the disjoint pair $(t{c}^{-1})\vee 1$ and $(c{t}^{-1})\vee 1=d$ gives the same contradiction. No normality assumption is used in this separation argument.