The Lagrange spectrum is contained in the Markov spectrum

For every real $t$, suppose there exists an irrational real $\xi$ with the following property: writing $D(x)=\inf\{|x-p|:p\in\mathbb Z\}$ and $u_n=1/((n+1)D((n+1)\xi))$ for $n\in\mathbb N=\{0,1,2,\ldots\}$, for every real $\varepsilon>0$ there is $N\in\mathbb N$ such that $u_n\le t+\varepsilon$ for every $n\ge N$, and for every real $\varepsilon>0$ and every $N\in\mathbb N$ there is $n\ge N$ such that $t-\varepsilon<u_n$. Then there exist real coefficients $A,B,C$ such that $\Delta=B^2-4AC>0$, $m=\inf\{|Ap^2+Bpq+Cq^2|:(p,q)\in\mathbb Z^2,\ p\ne0\text{ or }q\ne0\}>0$, and $t=\sqrt{\Delta}/m$, with the nonnegative square root. The infima defining $D$ and $m$ are over nonempty sets bounded below by zero; the latter uses every nonzero integer pair without requiring coprimality or attainment. The approximation sequence uses all positive natural denominators starting at one, so no zero denominator index is sampled, and irrationality makes each displayed approximation denominator positive. The limsup value $t$ is finite and real, the eventual upper inequality is non-strict, and the recurring lower inequality is strict. This is only the stated implication from the existence of an irrational witness to the existence of a coefficient triple.