[proofplan] We use the positive definiteness of the Cartan quadratic form attached to a finite Dynkin diagram. On each of the three arms, we define a real-valued test vector whose values increase linearly from the far endpoint toward a common value at the branch vertex. Evaluating the Cartan quadratic form on this vector reduces the desired inequality to the positivity of that quadratic form. [/proofplan] [step:Define the Cartan quadratic form and the linear test vector] Let $V$ be the vertex set of $\Gamma$, let $E$ be its edge set, and let $A = (A_{uv})_{u,v \in V}$ be the simply laced Cartan matrix associated to $\Gamma$, defined by \begin{align*} A_{uu} &= 2, \\ A_{uv} &= -1 \quad \text{if } \{u,v\} \in E, \\ A_{uv} &= 0 \quad \text{if } u \ne v \text{ and } \{u,v\} \notin E. \end{align*} Since $\Gamma$ is a finite Dynkin diagram, $A$ is positive definite. Hence the quadratic form \begin{align*} Q: \mathbb{R}^{V} &\to \mathbb{R} \\ x &\mapsto \frac{1}{2}\sum_{u,v \in V} A_{uv}x_u x_v \end{align*} satisfies $Q(x)>0$ for every nonzero $x \in \mathbb{R}^{V}$. Label the three arms by $a$, $b$, and $c$. On the arm of length $p$, write its vertices as \begin{align*} u_1,u_2,\dots,u_p, \end{align*} where $u_1$ is the endpoint and $u_p=v_0$. Similarly write the arms of lengths $q$ and $r$ as \begin{align*} w_1,\dots,w_q \quad \text{and} \quad z_1,\dots,z_r, \end{align*} with $w_q=z_r=v_0$. Define $x \in \mathbb{R}^{V}$ by \begin{align*} x_{u_i} &= \frac{i}{p}, && 1 \le i \le p, \\ x_{w_j} &= \frac{j}{q}, && 1 \le j \le q, \\ x_{z_k} &= \frac{k}{r}, && 1 \le k \le r. \end{align*} These definitions agree at the branch vertex because \begin{align*} x_{v_0}=\frac{p}{p}=\frac{q}{q}=\frac{r}{r}=1. \end{align*} Thus $x$ is a well-defined nonzero vector in $\mathbb{R}^{V}$. [/step] [step:Rewrite the quadratic form arm by arm] For a simply laced graph, the definition of $Q$ gives \begin{align*} Q(x) &= \sum_{v \in V} x_v^2 - \sum_{\{u,v\}\in E} x_u x_v. \end{align*} We compute the contribution from the arm of length $p$. Its vertices are $u_1,\dots,u_p$, and its edges are $\{u_i,u_{i+1}\}$ for $1 \le i \le p-1$. The terms involving this arm, excluding the branch contribution $x_{v_0}^2$ which will be counted once globally, are \begin{align*} \sum_{i=1}^{p-1} x_{u_i}^2 - \sum_{i=1}^{p-1} x_{u_i}x_{u_{i+1}}. \end{align*} Substituting $x_{u_i}=i/p$ gives \begin{align*} \sum_{i=1}^{p-1} \frac{i^2}{p^2} - \sum_{i=1}^{p-1} \frac{i(i+1)}{p^2} &= \frac{1}{p^2}\sum_{i=1}^{p-1}\bigl(i^2-i(i+1)\bigr) \\ &= -\frac{1}{p^2}\sum_{i=1}^{p-1} i \\ &= -\frac{p-1}{2p}. \end{align*} The same computation on the arms of lengths $q$ and $r$ gives contributions \begin{align*} -\frac{q-1}{2q} \quad \text{and} \quad -\frac{r-1}{2r}. \end{align*} Since the branch vertex contributes $x_{v_0}^2=1$, we obtain \begin{align*} Q(x) &= 1-\frac{p-1}{2p}-\frac{q-1}{2q}-\frac{r-1}{2r} \\ &= \frac{1}{2}\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}-1\right). \end{align*} [guided] The quadratic form attached to the simply laced Cartan matrix is \begin{align*} Q(x)=\sum_{v \in V}x_v^2-\sum_{\{u,v\}\in E}x_u x_v. \end{align*} This formula is useful because $\Gamma$ is a tree made from three paths meeting at one branch vertex, so we can evaluate the expression path by path. Consider first the arm of length $p$. Its vertices are $u_1,\dots,u_p$, with $u_p=v_0$, and the chosen vector has value $x_{u_i}=i/p$. We separate the branch term $x_{v_0}^2$ because the branch vertex belongs to all three arms but appears only once in the global sum $\sum_{v \in V}x_v^2$. The non-branch vertex terms and the edge terms along this arm give \begin{align*} \sum_{i=1}^{p-1} x_{u_i}^2 - \sum_{i=1}^{p-1}x_{u_i}x_{u_{i+1}} &= \sum_{i=1}^{p-1} \frac{i^2}{p^2} - \sum_{i=1}^{p-1} \frac{i(i+1)}{p^2} \\ &= \frac{1}{p^2}\sum_{i=1}^{p-1}\bigl(i^2-i(i+1)\bigr) \\ &= -\frac{1}{p^2}\sum_{i=1}^{p-1}i \\ &= -\frac{1}{p^2}\cdot \frac{(p-1)p}{2} \\ &= -\frac{p-1}{2p}. \end{align*} The arms of lengths $q$ and $r$ have the same linear pattern, so the identical computation gives \begin{align*} -\frac{q-1}{2q} \quad \text{and} \quad -\frac{r-1}{2r}. \end{align*} Adding the single branch contribution $x_{v_0}^2=1$ gives \begin{align*} Q(x) &= 1-\frac{p-1}{2p}-\frac{q-1}{2q}-\frac{r-1}{2r} \\ &= 1-\frac{1}{2}\left(3-\frac{1}{p}-\frac{1}{q}-\frac{1}{r}\right) \\ &= \frac{1}{2}\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}-1\right). \end{align*} This is the key computation: the sign of the Cartan square of the test vector is exactly the sign of the branch length expression. [/guided] [/step] [step:Use positive definiteness to force the strict inequality] The vector $x$ is nonzero because $x_{v_0}=1$. Since $A$ is positive definite, the quadratic form $Q$ satisfies $Q(x)>0$. From the computation above, \begin{align*} 0<Q(x)=\frac{1}{2}\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}-1\right). \end{align*} Multiplying by $2$ gives \begin{align*} \frac{1}{p}+\frac{1}{q}+\frac{1}{r}-1>0, \end{align*} and therefore \begin{align*} \frac{1}{p}+\frac{1}{q}+\frac{1}{r}>1. \end{align*} This proves the claimed branch length inequality. [/step]