China Mathematical Competition

Denote $s = |\vec{a}|$, $t = |\vec{b}|$, and then $s, t > 0$. Note that the included angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{3}$, and we have
$$\begin{aligned} |\vec{a} + \vec{b}|^2 &= \vec{a}^2 + (\vec{b})^2 + 2\vec{a} \cdot \vec{b} \\ &= s^2 + t^2 + 2st \cos \frac{\pi}{3} \\ &= s^2 + t^2 + st. \end{aligned}$$
By the condition, we know that $s, t, \sqrt{s^2 + t^2 + st}$ form arithmetic sequence in order. Then there is $\sqrt{s^2 + t^2 + st} = 2t - s$. Squaring and arranging the above equation gives $5st - 3t^2 = 0$. And since $t \neq 0$, it follows that $5s = 3t$, i.e., $s : t = 3 : 5$.

Therefore, $|\vec{a}| : |\vec{b}| : |\vec{a} + \vec{b}| = 3 : 5 : 7$.

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