[proofplan] This is the angle form of Toponogov comparison, proved from the previously established hinge form of Toponogov comparison as an external input. At a chosen vertex, the two minimizing sides issuing from that vertex form a nondegenerate geodesic hinge because the three vertices are pairwise distinct. If the angle in $M$ were smaller than the model angle, the constant-curvature law of cosines would make the corresponding model opposite side strictly smaller than the actual opposite side, contradicting the hinge comparison. Repeating the same verified hinge argument at the three vertices gives the three asserted inequalities. [/proofplan]

[step:Fix the notation for the three geodesic hinges] Since $p$, $q$, and $r$ are pairwise distinct, the side lengths satisfy \begin{align*} a>0, \qquad b>0, \qquad c>0. \end{align*} Let \begin{align*} \gamma_{pq}: [0,c] &\to M, & \gamma_{pr}: [0,b] &\to M, & \gamma_{qr}: [0,a] &\to M \end{align*} be the chosen unit-speed minimizing geodesic segments satisfying \begin{align*} \gamma_{pq}(0)=p,\quad \gamma_{pq}(c)=q,\qquad \gamma_{pr}(0)=p,\quad \gamma_{pr}(b)=r,\qquad \gamma_{qr}(0)=q,\quad \gamma_{qr}(a)=r. \end{align*} The angle at $p$ is the Riemannian angle between $\gamma_{pq}'(0)$ and $\gamma_{pr}'(0)$. The angle at $q$ is the angle between $-\gamma_{pq}'(c)$ and $\gamma_{qr}'(0)$. The angle at $r$ is the angle between $-\gamma_{pr}'(b)$ and $-\gamma_{qr}'(a)$. Let \begin{align*} \alpha := \angle qpr,\qquad \beta := \angle pqr,\qquad \rho := \angle prq \end{align*} denote the angles of the chosen triangle in $M$, and let \begin{align*} \bar{\alpha} := \angle \bar{q}\bar{p}\bar{r},\qquad \bar{\beta} := \angle \bar{p}\bar{q}\bar{r},\qquad \bar{\rho} := \angle \bar{p}\bar{r}\bar{q} \end{align*} denote the corresponding model angles in $M_k^2$. [/step]

[step:Record the monotonicity of the model opposite side]In the model space $M_k^2$, fix two side lengths $x,y>0$ for which the corresponding model hinges are defined. Let \begin{align*} s_k^{x,y}: [0,\pi] &\to [0,\infty) \end{align*} denote the function assigning to an angle $\theta$ the length of the side opposite a model hinge with adjacent side lengths $x$ and $y$ and included angle $\theta$. On the range of angles relevant to the given model triangle, this function is strictly increasing. Indeed, in the Euclidean case $k=0$, the model law of cosines gives \begin{align*} \bigl(s_0^{x,y}(\theta)\bigr)^2 = x^2+y^2-2xy\cos \theta, \end{align*} which is strictly increasing in $\theta$ because $\cos \theta$ is strictly decreasing on $[0,\pi]$ and $x,y>0$. For $k>0$, the spherical law of cosines gives \begin{align*} \cos(\sqrt{k}\, s_k^{x,y}(\theta)) = \cos(\sqrt{k}\,x)\cos(\sqrt{k}\,y) + \sin(\sqrt{k}\,x)\sin(\sqrt{k}\,y)\cos\theta. \end{align*} The assumptions on the side lengths and perimeter ensure that the relevant side lengths lie in $(0,\pi/\sqrt{k})$, so the function $t \mapsto \cos(\sqrt{k}\,t)$ is strictly decreasing on that interval. Since $\sin(\sqrt{k}\,x)>0$, $\sin(\sqrt{k}\,y)>0$, and $\theta \mapsto \cos\theta$ is strictly decreasing on $[0,\pi]$, the opposite side length $s_k^{x,y}(\theta)$ is strictly increasing in $\theta$. For $k<0$, the hyperbolic law of cosines gives \begin{align*} \cosh(\sqrt{-k}\, s_k^{x,y}(\theta)) = \cosh(\sqrt{-k}\,x)\cosh(\sqrt{-k}\,y) - \sinh(\sqrt{-k}\,x)\sinh(\sqrt{-k}\,y)\cos\theta. \end{align*} Since $t \mapsto \cosh(\sqrt{-k}\,t)$ is strictly increasing on $[0,\infty)$, $\sinh(\sqrt{-k}\,x)>0$, $\sinh(\sqrt{-k}\,y)>0$, and $\theta \mapsto -\cos\theta$ is strictly increasing on $[0,\pi]$, the opposite side length $s_k^{x,y}(\theta)$ is strictly increasing in $\theta$.[/step]

[guided]The point of this step is to justify the only algebraic fact about model triangles that we will use: if the two sides of a model hinge are fixed, then increasing the included angle increases the opposite side. Define \begin{align*} s_k^{x,y}: [0,\pi] &\to [0,\infty) \end{align*} as follows: $s_k^{x,y}(\theta)$ is the length of the side opposite a model hinge in $M_k^2$ whose adjacent side lengths are $x$ and $y$ and whose included angle is $\theta$. We prove that this function is strictly increasing on the range where the model hinge is defined. When $k=0$, the model space is Euclidean. The Euclidean law of cosines gives \begin{align*} \bigl(s_0^{x,y}(\theta)\bigr)^2 = x^2+y^2-2xy\cos \theta. \end{align*} Here $x>0$ and $y>0$, so the coefficient $2xy$ is positive. Since $\theta \mapsto \cos\theta$ is strictly decreasing on $[0,\pi]$, the quantity $x^2+y^2-2xy\cos\theta$ is strictly increasing in $\theta$. Taking the non-negative square root preserves strict increase, so $s_0^{x,y}$ is strictly increasing. When $k>0$, the model space is the sphere of radius $1/\sqrt{k}$. The spherical law of cosines gives \begin{align*} \cos(\sqrt{k}\, s_k^{x,y}(\theta)) = \cos(\sqrt{k}\,x)\cos(\sqrt{k}\,y) + \sin(\sqrt{k}\,x)\sin(\sqrt{k}\,y)\cos\theta. \end{align*} The side length assumptions ensure that the relevant values of $x$, $y$, and $s_k^{x,y}(\theta)$ lie in $(0,\pi/\sqrt{k})$. Therefore $\sin(\sqrt{k}\,x)>0$ and $\sin(\sqrt{k}\,y)>0$, while the function $t \mapsto \cos(\sqrt{k}\,t)$ is strictly decreasing on $(0,\pi/\sqrt{k})$. As $\theta$ increases, $\cos\theta$ strictly decreases, so the right-hand side strictly decreases. Because the left-hand side is a strictly decreasing function of the side length, the side length itself must strictly increase. When $k<0$, the model space is hyperbolic. The hyperbolic law of cosines gives \begin{align*} \cosh(\sqrt{-k}\, s_k^{x,y}(\theta)) = \cosh(\sqrt{-k}\,x)\cosh(\sqrt{-k}\,y) - \sinh(\sqrt{-k}\,x)\sinh(\sqrt{-k}\,y)\cos\theta. \end{align*} Since $x,y>0$, the factor $\sinh(\sqrt{-k}\,x)\sinh(\sqrt{-k}\,y)$ is positive. Thus the right-hand side is strictly increasing as $\theta$ increases. The function $t \mapsto \cosh(\sqrt{-k}\,t)$ is strictly increasing on $[0,\infty)$, so $s_k^{x,y}(\theta)$ is strictly increasing.[/guided]

[step:Compare the angle at $p$ by contradiction]We prove $\alpha \geq \bar{\alpha}$. Suppose, for contradiction, that $\alpha < \bar{\alpha}$. Consider the geodesic hinge in $M$ formed by the two unit-speed minimizing geodesics \begin{align*} \gamma_{pq}: [0,c] &\to M, & \gamma_{pr}: [0,b] &\to M \end{align*} issuing from $p$. Its side lengths adjacent to the angle are $c$ and $b$, its included angle is $\alpha$, and its opposite endpoint distance is \begin{align*} d_M(\gamma_{pq}(c),\gamma_{pr}(b)) = d_M(q,r)=a. \end{align*} We verify the hypotheses of the hinge comparison input. The manifold $(M,g)$ is complete and satisfies $\sec_M \geq k$ by hypothesis. The curves $\gamma_{pq}$ and $\gamma_{pr}$ are unit-speed minimizing geodesic segments by construction. Their lengths $c$ and $b$ are positive because the vertices are pairwise distinct. When $k>0$, the required length restrictions hold because $b<\pi/\sqrt{k}$ and $c<\pi/\sqrt{k}$ are among the stated assumptions, and the model opposite length $a$ also satisfies $a<\pi/\sqrt{k}$. Therefore the hinge form of the Toponogov comparison theorem applies to this hinge: the model hinge in $M_k^2$ with adjacent side lengths $c$ and $b$ and included angle $\alpha$ has opposite side length $a_\alpha$ satisfying \begin{align*} a \leq a_\alpha. \end{align*} On the other hand, the given model triangle $\bar{p}\bar{q}\bar{r}$ has adjacent side lengths $c$ and $b$ at $\bar{p}$, included angle $\bar{\alpha}$, and opposite side length $a$. By the strict monotonicity of $s_k^{c,b}$, \begin{align*} a_\alpha = s_k^{c,b}(\alpha) < s_k^{c,b}(\bar{\alpha}) = a. \end{align*} This contradicts $a \leq a_\alpha$. Hence $\alpha \geq \bar{\alpha}$.[/step]

[guided]The contradiction argument depends on a precise hinge comparison. We first check each hypothesis before using it. The ambient manifold $(M,g)$ is complete, and its sectional curvature satisfies $\sec_M \geq k$; these are hypotheses of the theorem. The two sides of the hinge are the maps \begin{align*} \gamma_{pq}: [0,c] &\to M, & \gamma_{pr}: [0,b] &\to M, \end{align*} and they are unit-speed minimizing geodesic segments by the chosen-triangle hypothesis. The side lengths $c$ and $b$ are positive because $p$, $q$, and $r$ are pairwise distinct, so the angle $\alpha$ is a genuine hinge angle. If $k>0$, the spherical version of the hinge comparison requires the relevant side lengths to lie below $\pi/\sqrt{k}$; here this follows from \begin{align*} a<\frac{\pi}{\sqrt{k}}, \qquad b<\frac{\pi}{\sqrt{k}}, \qquad c<\frac{\pi}{\sqrt{k}}. \end{align*} Thus the hinge form of Toponogov comparison applies. It says that the distance between the two endpoints in $M$ is no larger than the opposite side of the model hinge with the same adjacent side lengths and the same included angle. Since those endpoints are $q$ and $r$, this gives \begin{align*} a=d_M(q,r)\leq a_\alpha, \end{align*} where $a_\alpha$ denotes the opposite side length of the model hinge in $M_k^2$ with adjacent side lengths $c$ and $b$ and included angle $\alpha$.[/guided]

[step:Compare the angles at $q$ and $r$ in the same way] We next prove $\beta \geq \bar{\beta}$. Suppose, for contradiction, that $\beta < \bar{\beta}$. The geodesic hinge in $M$ based at $q$ has adjacent side lengths \begin{align*} d_M(q,p)=c,\qquad d_M(q,r)=a, \end{align*} included angle $\beta$, and opposite endpoint distance \begin{align*} d_M(p,r)=b. \end{align*} The hinge comparison hypotheses are satisfied: $(M,g)$ is complete, $\sec_M\geq k$, the two sides from $q$ are minimizing geodesic segments, and their lengths $c$ and $a$ are positive. When $k>0$, the required length restrictions follow from $a<\pi/\sqrt{k}$, $b<\pi/\sqrt{k}$, and $c<\pi/\sqrt{k}$. Hence the hinge form of Toponogov comparison gives that the model hinge with adjacent side lengths $c$ and $a$ and included angle $\beta$ has opposite side length $b_\beta$ satisfying \begin{align*} b \leq b_\beta. \end{align*} The model triangle $\bar{p}\bar{q}\bar{r}$ has adjacent side lengths $c$ and $a$ at $\bar{q}$, included angle $\bar{\beta}$, and opposite side length $b$. Since $s_k^{c,a}$ is strictly increasing, \begin{align*} b_\beta = s_k^{c,a}(\beta) < s_k^{c,a}(\bar{\beta}) = b, \end{align*} contradicting $b \leq b_\beta$. Therefore $\beta \geq \bar{\beta}$. Finally, we prove $\rho \geq \bar{\rho}$. Suppose, for contradiction, that $\rho < \bar{\rho}$. The geodesic hinge in $M$ based at $r$ has adjacent side lengths \begin{align*} d_M(r,p)=b,\qquad d_M(r,q)=a, \end{align*} included angle $\rho$, and opposite endpoint distance \begin{align*} d_M(p,q)=c. \end{align*} The hinge comparison hypotheses are satisfied again: $(M,g)$ is complete, $\sec_M\geq k$, the two sides from $r$ are minimizing geodesic segments, and their lengths $b$ and $a$ are positive. When $k>0$, the required length restrictions follow from $a<\pi/\sqrt{k}$, $b<\pi/\sqrt{k}$, and $c<\pi/\sqrt{k}$. Hence the hinge form of Toponogov comparison gives that the model hinge with adjacent side lengths $b$ and $a$ and included angle $\rho$ has opposite side length $c_\rho$ satisfying \begin{align*} c \leq c_\rho. \end{align*} The model triangle $\bar{p}\bar{q}\bar{r}$ has adjacent side lengths $b$ and $a$ at $\bar{r}$, included angle $\bar{\rho}$, and opposite side length $c$. Since $s_k^{b,a}$ is strictly increasing, \begin{align*} c_\rho = s_k^{b,a}(\rho) < s_k^{b,a}(\bar{\rho}) = c, \end{align*} contradicting $c \leq c_\rho$. Therefore $\rho \geq \bar{\rho}$. [/step]

[step:Translate the notation back to the stated angle inequalities] By the definitions of $\alpha$, $\beta$, $\rho$ and of the model angles $\bar{\alpha}$, $\bar{\beta}$, $\bar{\rho}$, the inequalities proved above are exactly \begin{align*} \angle qpr &\geq \angle \bar{q}\bar{p}\bar{r}, \\ \angle pqr &\geq \angle \bar{p}\bar{q}\bar{r}, \\ \angle prq &\geq \angle \bar{p}\bar{r}\bar{q}. \end{align*} Reordering the last two lines gives the angle comparison in the theorem statement. This completes the proof. [/step]