I want to check the definition for the Weyl vector $\rho$ for $A_2$ in Dynkin basis.
Here $\rho := \frac{1}{2}\sum_{\alpha>0} \alpha = \sum_{i=1}^r \Lambda_{(i)}$. Here $\alpha > 0$ are the positive roots of $A_2$ and $\Lambda_{(i)}$ are the elements of the basis of the weight space (In my book this is defined as the Dynkin basis, apparently this is not the most common name though). They are also called the fundamental weights.
I have no idea on how to prove this. I started by looking at the Dynkin diagram, but it seemed a lot of work to first calculate the roots...
The Cartan matrix of $A_2=\mathfrak{sl}(3,\mathbb{R})$ is $\begin{bmatrix} 2&-1\\-1&2 \end{bmatrix}$ which is obtainable from the Dynkin diagram. Then $\Delta=\{\alpha_1,\alpha_2\}$ is a basis of the root system, and $\left\{ \dfrac{2\alpha_1}{(\alpha_1,\alpha_1)},\dfrac{2\alpha_2}{(\alpha_2,\alpha_2)} \right\}$ form a basis of the Euclidean plane $\mathbb{R}^2$. Let $\{\Lambda_1,\Lambda_2\}$ be the dual basis relative to the inner product on $\mathbb{R}^2,$ i.e. $$ \dfrac{2(\Lambda_i,\alpha_j)}{(\alpha_j,\alpha_j)}=\delta_{ij} $$ Back to the Cartan matrix, we get $\alpha_1=2\Lambda_1-\Lambda_2$ and $\alpha_2=\Lambda_2-2\Lambda_1.$ The inverse matrix is $\dfrac{1}{3}\begin{bmatrix} 2&1\\1&2 \end{bmatrix},$ hence $\Lambda_1=\frac{1}{3}(2\alpha_1+\alpha_2),\Lambda_2=\frac{1}{3}(\alpha_1+2\alpha_2).$
Addition gives you the desired result; don't forget that $\alpha_1+\alpha_2$ is also a positive root.