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Let $M$ be a Hadamard manifold. Let $\Delta(p_1,p_2,p_3)$ be a geodesic triangle and for $i=1,2,3\pmod 3$ , $\gamma_i:[0,\ell_i]\to M$ be the geodesic joining $p_i$ to $p_{i+1}$ where $\ell_i=\ell(\gamma_i)$. If $\alpha_i$ is the angle between $\dot\gamma_i(0)$ and $-\dot\gamma_{i-1}(\ell_{i-1})$, then how to prove that $$\alpha_1+\alpha_2+\alpha_3\le\pi\tag{1}$$ and $$\ell_i^2+\ell_{i+1}^2-2\ell_i\ell_{i+1}\cos(\alpha_{i+1})\le\ell_{i-1}^2\tag{2}$$ I'm looked some books on Riemannian geometry but did not find a clear proof. Some references introduce this as a result of Rauch's comparison theorem but I don't have enough information and experience on Riemannian geometry, while I can understand Rauch's comparison theorem's statement but I don't know how to obtain $(1)$ and $(2)$ from this theorem, so it would be appropriate if someone here guide me and sketch a proof.

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We need an application of Rauch's comparison theorem, indeed let $M$ and $N$ be two Riemannian manifolds with dimensions $m$ and $n$, respectively such that $n\ge m$ and $K_M\le K_N$. For $p\in M$ and $q\in N$ suppose $\epsilon>0$ is sufficiently small such that $\exp_p|_{B(0;\epsilon)}$ is a diffeomorphism and $\exp_q|_{B(0;\epsilon)}$ is non-singular. If $\gamma:[0,a]\to\exp_p(B(0;\epsilon))$ is a smooth path and $\lambda:[0,a]\to\exp_q(B(0;\epsilon))$ for all $s\in[0,a]$ is defined by $$\lambda(s)=(\exp_q\circ i\circ\exp^{-1}_p)(\gamma(s))$$ where $i:T_pM\to T_qN$ is linear isometry, then $\ell(\gamma)\ge\ell(\lambda)$.

Denote the geodesics of geodesic triangle by $\gamma_A$, $\gamma_B$ and $\gamma_C$, where the subscripts are the length of the geodesic segments. Let $\Gamma_i(t)=\exp^{-1}_c({\gamma_i})$ for $i=A,B,C$. These are curves in $T_cM$ and since $\gamma_A$ and $\gamma_B$ are radial geodesics, we have that $A=\ell(\gamma_A)=\ell(\Gamma_A)$ and $B=\ell(\gamma_B)=\ell(\Gamma_B)$. Furthermore, if you join the two endpoints of $\Gamma_C$ by a straight line $\Gamma$, then by the mentioned application $\ell(\Gamma)\le\ell(\Gamma_C)\le\ell(\gamma_C)$. Thus

$$A^2+B^2-2AB\cos\alpha_3=\ell^2(\Gamma)\le\ell^2(\Gamma_C)\le\ell^2(\gamma_C)=C^2,$$

Now, consider a triangle with side lengths $A$, $B$ and $C$ in Euclidean space, and denote interior angles $\alpha'_1$, $\alpha'_2$ and $\alpha'_3$. Observe that

$$A^2+B^2-2AB\cos\alpha_3\le C^2=A^2+B^2-2AB\cos\alpha'_3,$$

so that $\cos\alpha_3\ge\cos\alpha'_3$. Since cosine is a strictly decreasing function on $[0,\pi]$ so $\alpha_3\le\alpha'_3$. Repeating this for the other two angles yields $$\alpha_1+\alpha_2+\alpha_3\le\pi.$$

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I assume by Hadamard manifold, you mean nonpositive curvature and simply connected. Then the second statement follows from the fact that the exponential map a one of the vertices of your triangle is distance nondecreasing, which follows from the Rauch comparison theorem and the law of cosines in the tangent space.

The first statement follows similarly. Namely, open the hinge in the hinge in the tangent space until the distance between the points in the tangent space is equal to the length of the third side of the downstairs. Then the triangle in the tangent space has three sides equal to the triangle downstairs and one angle bigger (than or equal to). But the Euclidean triangle is determined up to congruence by its side lengths so ${\color{red}{\text{all}}}$ of its angles must be at least as big as those in the Hadamard manifold (and in the equality case, the triangle in the Hadamard manifold must span a flat $2$-plane).

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