[proofplan] The proof is obtained by taking the trace of the matrix Riccati equation on the $(n-1)$-dimensional space orthogonal to the radial direction. The trace of the radial curvature endomorphism is exactly the Ricci curvature in the radial direction. The only estimate needed is the finite-dimensional inequality \begin{align*} \operatorname{tr}(A^2)\geq \frac{(\operatorname{tr}A)^2}{n-1} \end{align*} for a self-adjoint endomorphism $A$. [/proofplan]

[step:Take the trace of the radial matrix Riccati equation] Fix $t \in (0,a)$. Choose a $g_{\gamma(t)}$-[orthonormal basis](/page/Orthonormal%20Basis) $(e_1,\dots,e_{n-1})$ of $E_t$. Let $I \subset (0,a)$ be an open interval containing $t$ on which the basis is extended by parallel transport to a $g_{\gamma(s)}$-orthonormal frame of $E_s$ along $\gamma$. Thus, for each $1 \leq i \leq n-1$, we have a smooth vector field \begin{align*} E_i: I &\to TM \\ s &\mapsto E_i(s) \in E_s \end{align*} with $E_i(t)=e_i$ and $\nabla_{\dot\gamma}E_i=0$ on $I$. Taking the trace over $E_t$ in \begin{align*} A'(t)+A(t)^2+R_\gamma(t)=0 \end{align*} gives \begin{align*} \operatorname{tr}_{E_t} A'(t)+\operatorname{tr}_{E_t}(A(t)^2)+\operatorname{tr}_{E_t} R_\gamma(t)=0. \end{align*} For $s \in I$, the trace of $A(s)$ in the parallel orthonormal frame is \begin{align*} m(s)=\operatorname{tr}_{E_s}A(s)=\sum_{i=1}^{n-1}g_{\gamma(s)}(A(s)E_i(s),E_i(s)). \end{align*} Differentiating at $s=t$, using metric compatibility of the Levi-Civita connection and $\nabla_{\dot\gamma}E_i=0$, gives \begin{align*} m'(t) &=\sum_{i=1}^{n-1}g_{\gamma(t)}(A'(t)e_i,e_i)\\ &=\operatorname{tr}_{E_t}A'(t). \end{align*} Moreover, since $(e_1,\dots,e_{n-1},\dot{\gamma}(t))$ is a $g_{\gamma(t)}$-orthonormal basis of $T_{\gamma(t)}M$ and the curvature term in the $\dot{\gamma}(t)$ direction vanishes by the alternating symmetry of the curvature tensor, the trace of $R_\gamma(t)$ is \begin{align*} \operatorname{tr}_{E_t}R_\gamma(t) &=\sum_{i=1}^{n-1} g_{\gamma(t)}(R(e_i,\dot{\gamma}(t))\dot{\gamma}(t),e_i)\\ &=\operatorname{Ric}_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t)). \end{align*} Therefore \begin{align*} m'(t)+\operatorname{tr}_{E_t}(A(t)^2)+\operatorname{Ric}_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t))=0. \end{align*} [/step]

[step:Bound the quadratic trace term by the square of the mean curvature]Let $(e_1,\dots,e_{n-1})$ be a $g_{\gamma(t)}$-orthonormal basis of $E_t$, and define the matrix coefficients \begin{align*} a_{ij}:=g_{\gamma(t)}(A(t)e_j,e_i) \end{align*} for $1 \leq i,j \leq n-1$. Since $A(t)$ is self-adjoint, the matrix $(a_{ij})$ is symmetric. Hence \begin{align*} \operatorname{tr}_{E_t}(A(t)^2) &=\sum_{i=1}^{n-1} g_{\gamma(t)}(A(t)^2e_i,e_i)\\ &=\sum_{i=1}^{n-1}\sum_{j=1}^{n-1} a_{ij}a_{ji}\\ &=\sum_{i=1}^{n-1}\sum_{j=1}^{n-1} a_{ij}^2\\ &\geq \sum_{i=1}^{n-1} a_{ii}^2. \end{align*} Applying the [Cauchy-Schwarz Inequality](/page/Cauchy-Schwarz%20Inequality) in $\mathbb{R}^{n-1}$ to the vectors $(a_{11},\dots,a_{n-1,n-1})$ and $(1,\dots,1)$ gives \begin{align*} \left(\sum_{i=1}^{n-1}a_{ii}\right)^2 \leq \left(\sum_{i=1}^{n-1}a_{ii}^2\right)(n-1). \end{align*} Since \begin{align*} m(t)=\operatorname{tr}_{E_t}A(t)=\sum_{i=1}^{n-1}a_{ii}, \end{align*} we obtain \begin{align*} \operatorname{tr}_{E_t}(A(t)^2)\geq \frac{m(t)^2}{n-1}. \end{align*}[/step]

[guided]The goal of this step is to replace the full quadratic quantity $\operatorname{tr}_{E_t}(A(t)^2)$ by an expression involving only the scalar trace $m(t)$. This is possible because $A(t)$ is self-adjoint on the $(n-1)$-dimensional [inner product space](/page/Inner%20Product%20Space) $E_t$. Choose a $g_{\gamma(t)}$-orthonormal basis $(e_1,\dots,e_{n-1})$ of $E_t$. Define the [real numbers](/page/Real%20Numbers) \begin{align*} a_{ij}:=g_{\gamma(t)}(A(t)e_j,e_i) \end{align*} for $1 \leq i,j \leq n-1$. These are the matrix coefficients of $A(t)$ in the chosen orthonormal basis. Since $A(t)$ is self-adjoint, we have $a_{ij}=a_{ji}$ for all $i,j$. Therefore \begin{align*} \operatorname{tr}_{E_t}(A(t)^2) &=\sum_{i=1}^{n-1} g_{\gamma(t)}(A(t)^2e_i,e_i)\\ &=\sum_{i=1}^{n-1}\sum_{j=1}^{n-1} a_{ij}a_{ji}\\ &=\sum_{i=1}^{n-1}\sum_{j=1}^{n-1} a_{ij}^2. \end{align*} This expression is a sum of squares. Discarding the off-diagonal non-negative terms gives \begin{align*} \operatorname{tr}_{E_t}(A(t)^2) \geq \sum_{i=1}^{n-1}a_{ii}^2. \end{align*} Now we compare the sum of diagonal squares with the square of the sum of diagonal entries. Apply the [Cauchy-Schwarz Inequality](/page/Cauchy-Schwarz%20Inequality) in the Euclidean space $\mathbb{R}^{n-1}$ to the vectors \begin{align*} (a_{11},\dots,a_{n-1,n-1}) \quad\text{and}\quad (1,\dots,1). \end{align*} It gives \begin{align*} \left(\sum_{i=1}^{n-1}a_{ii}\right)^2 \leq \left(\sum_{i=1}^{n-1}a_{ii}^2\right) \left(\sum_{i=1}^{n-1}1^2\right) = (n-1)\sum_{i=1}^{n-1}a_{ii}^2. \end{align*} Rearranging, \begin{align*} \sum_{i=1}^{n-1}a_{ii}^2 \geq \frac{1}{n-1}\left(\sum_{i=1}^{n-1}a_{ii}\right)^2. \end{align*} Since the trace of $A(t)$ is the sum of its diagonal entries in any orthonormal basis, \begin{align*} m(t)=\operatorname{tr}_{E_t}A(t)=\sum_{i=1}^{n-1}a_{ii}. \end{align*} Combining the preceding inequalities yields \begin{align*} \operatorname{tr}_{E_t}(A(t)^2)\geq \frac{m(t)^2}{n-1}. \end{align*}[/guided]

[step:Substitute the trace bound into the traced Riccati equation] From the traced Riccati equation, \begin{align*} m'(t)+\operatorname{tr}_{E_t}(A(t)^2)+\operatorname{Ric}_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t))=0. \end{align*} Using \begin{align*} \operatorname{tr}_{E_t}(A(t)^2)\geq \frac{m(t)^2}{n-1}, \end{align*} we obtain \begin{align*} m'(t)+\frac{m(t)^2}{n-1}+\operatorname{Ric}_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t)) \leq m'(t)+\operatorname{tr}_{E_t}(A(t)^2)+\operatorname{Ric}_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t)) =0. \end{align*} Thus, for every $t \in (0,a)$, \begin{align*} m'(t)+\frac{m(t)^2}{n-1}+\operatorname{Ric}_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t))\leq 0. \end{align*} This is the desired trace Riccati inequality. [/step]