Let $(X,d)$ be a geodesic space, suppose $g:X \to \mathbb{R}$ is geodesically $\lambda$ convex with $\lambda >0$, that is for all $x,y \in X$ there exists a constant speed geodesic $\gamma$ from $x$ to $y$ such that for all $s\in [0,1]$ $$ g(\gamma_s) \leq s g(x) + (1-s) g(y) -\frac{\lambda}{2}s(1-s)d(x,y)^2 $$ Let $x_0$ be the unique minimizer of $g$. Show that $\forall x$ $$ g(x) \geq g(x_0) + \frac{\lambda}{2}d(x,x_0)^2 $$
EDIT: thanks for the comment. Honestly I did not do much because I was stuck from the beginning, since I had no idea how to use the geodesic property. I just picked a geodesic between $x$ and $x_0$ and tried to plug in the property but did not see how that could help! Any hint or suggestions?
Choose your geodesic with $\gamma_{0}=x_{0}$ and $\gamma_{1}=x$. Since $x_{0}$ is minimal we have \begin{equation} 0=\frac{d}{ds}|_{s=0}g(\gamma_{s}). \end{equation} Then, using your first inequality it follows \begin{equation} 0=\lim_{h\to 0}\frac{1}{h}(g(\gamma_{h})-g(x_{0}))\le\lim_{h\to 0}\frac{1}{h}\left(hg(x)+(1-h)g(x_{0})-\frac{\lambda}{2}h(1-h)d^{2}(x,x_{0})-g(x_{0})\right) \end{equation} which yields the claim.