Probability that a stick randomly broken in three places can form a triangle

The lengths of the broken stick pieces $(Y_1, Y_2, Y_3, Y_4)$ are spacings of ordered uniform. The joint distribution is equivalent to $$ \left(\frac {X_1} {\sum_{i=1}^4 X_i}, \frac {X_2} {\sum_{i=1}^4 X_i}, \frac {X_3} {\sum_{i=1}^4 X_i}, \frac {X_4} {\sum_{i=1}^4 X_i}\right)$$ where $X_i$ are iid exponential random variables. If we ordered the above spacings, the distribution of $(Y_{(1)}, Y_{(2)}, Y_{(3)}, Y_{(4)})$ is equivalent to

$$ \left(\frac {X_1/4} {\sum_{i=1}^4 X_i}, \frac {X_1/4 + X_2/3} {\sum_{i=1}^4 X_i}, \frac {X_1/4 + X_2/3 + X_3/2 } {\sum_{i=1}^4 X_i}, \frac {X_1/4 + X_2/3 + X_3/2 + X_4} {\sum_{i=1}^4 X_i}\right)$$

Any $3$ of the $4$ pieces cannot form a triangle if and only if $$ \begin{cases} Y_{(1)} + Y_{(2)} < Y_{(3)} \\ Y_{(1)} + Y_{(2)} < Y_{(4)} \\ Y_{(1)} + Y_{(3)} < Y_{(4)} \\ Y_{(2)} + Y_{(3)} < Y_{(4)} \end{cases}$$ Note that the second inequality is implied by the first, and the third inequality is implied by the fourth. So the probability of no triangle being formed is $$ \begin{align} &\Pr\{ Y_{(1)} + Y_{(2)} < Y_{(3)}, Y_{(2)} + Y_{(3)} < Y_{(4)}\} \\ =& \Pr\Bigg\{\frac {X_1} {4} + \frac {X_1} {4} + \frac {X_2} {3} < \frac {X_1} {4} + \frac {X_2} {3} + \frac {X_3} {2}, \\ & \frac {X_1} {4} + \frac {X_2} {3} + \frac {X_1} {4} + \frac {X_2} {3} + \frac {X_3} {2} < \frac {X_1} {4} + \frac {X_2} {3} + \frac {X_3} {2} + X_4\Bigg\} \\ =& \Pr\{X_1 < 2X_3, 3X_1 + 4X_2 < 12X_4\} \\ =& \int_0^{\infty} \Pr\{2X_3 > x\}\Pr\{12X_4 - 4X_2 > 3x\}e^{-x}dx \end{align}$$

Note that $$ \begin{align} &\Pr\{12X_4 - 4X_2 > 3x\} \\ =& \int_0^{\infty}\Pr\{12X_4 - 4u > 3x\}e^{-u}du \\ =& \int_0^{\infty} e^{-(3x+4u)/12} e^{-u}du \\ =& e^{-x/4}\int_0^{\infty} e^{-4u/3}du \\ =& \frac {3} {4}e^{-x/4} \end{align} $$

So the integral become $$ \int_0^{\infty} e^{-x/2}\frac {3} {4} e^{-x/4}e^{-x}dx = \frac {3} {4} \int_0^{\infty} e^{-7x/4} dx = \frac {3} {4} \times \frac {4} {7} = \frac {3} {7} $$

This is not a very elegant way but at least it is doable. Looking forward to someone to post a better solution.