= Solution
The original PDF has \b[absolute-value bars around both gradients]; they are missing from the transcription. The stability argument must respect the sign of the upward coordinate.
Displace a small parcel upward by $\xi$. Assume it remains in <pressure equilibrium> with its surroundings but exchanges negligible heat, so it follows an <adiabatic process>. For a perfect gas with <adiabatic index> $\gamma$, $T\propto p^{(\gamma-1)/\gamma}$ along that parcel's adiabat. Its <temperature> gradient is therefore
$$
T'_{\rm ad}=\frac{\gamma-1}{\gamma}\frac Tp\frac{dp}{dz}
=-\frac{g}{c_p},\qquad c_p=\frac{\gamma\mathcal R}{\gamma-1}.
$$
The last step uses <hydrostatic equilibrium>, and $c_p$ is the <specific heat capacity at constant pressure>. At its new height, the <temperature> difference between the parcel and the environment is $(T'_{\rm ad}-T')\xi$. At equal <pressure>, its relative density excess is the negative of its relative <temperature> excess. The upward <buoyancy> acceleration is thus
$$
\ddot\xi=\frac gT(T'_{\rm ad}-T')\xi=-N^2\xi,\qquad
N^2=\frac gT\left(\frac{dT}{dz}+\frac g{c_p}\right).
$$
This is the squared <Brunt–Väisälä frequency>. <Convective stability> requires $N^2>0$, or, in signed-gradient form,
$$
\boxed{\frac{dT}{dz}>\left(1-\frac1\gamma\right)\frac Tp\frac{dp}{dz}=-\frac g{c_p}.}
$$
For a <temperature> decreasing upward, this is exactly the PDF's <Schwarzschild criterion>
$$
\boxed{\left|\frac{dT}{dz}\right|<\left(1-\frac1\gamma\right)\frac Tp
\left|\frac{dp}{dz}\right|.}
$$
The absolute-value condition is also sufficient more generally, although a sufficiently steep <temperature> inversion remains stable even when that stronger condition fails. Reversing the signed inequality, as the bar-free transcription does, would instead describe an unstable <temperature> decrease.
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