6. Let non-negative real numbers $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ satisfy $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=1$, then the minimum value of $\frac{\alpha_{1}}{1+\alpha_{2}+\alpha_{3}+\cdots+\alpha_{n}}+$ $\frac{\alpha_{2}}{1+\alpha_{1}+\alpha_{3}+\cdots+\alpha_{n}}+\cdots+\frac{\alpha_{n}}{1+\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n-1}}$ is $\qquad$.
6. $\frac{n}{2 n-1}$ Detailed Explanation: For each fraction in the given expression, we pair it with a constant 1, then perform a common denominator operation. After that, we replace $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}$ with the constant 1, obtaining $\frac{\alpha_{1}}{1+\alpha_{2}+\alpha_{3}+\cdots+\alpha_{n}}+1=\frac{1+\left(\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}\right)}{2-\alpha_{1}}=\frac{2}{2-\alpha_{1}}$.
Similarly, we get $\frac{\alpha_{2}}{1+\alpha_{1}+\alpha_{3}+\cdots+\alpha_{n}}+1=\frac{2}{2-\alpha_{2}}, \cdots, \frac{\alpha_{n}}{1+\alpha_{1}+\cdots+\alpha_{n-1}}+1=\frac{2}{2-\alpha_{n}}$.
Let $y=\frac{\alpha_{1}}{1+\alpha_{2}+\alpha_{3}+\cdots+\alpha_{n}}+\frac{\alpha_{2}}{1+\alpha_{1}+\alpha_{3}+\cdots+\alpha_{n}}+\cdots+\frac{\alpha_{n}}{1+\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n-1}}$.
Then $y+n=\frac{2}{2-\alpha_{1}}+\frac{2}{2-\alpha_{2}}+\cdots+\frac{2}{2-\alpha_{n}}$. Note that
$\left(2-\alpha_{1}\right)+\left(2-\alpha_{2}\right)+\cdots+\left(2-\alpha_{n}\right)=2 n-\left(\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}\right)=2 n-1$.
Thus, $(2 n-1)\left(\frac{1}{2-\alpha_{1}}+\frac{1}{2-\alpha_{2}}+\cdots+\frac{1}{2-\alpha_{n}}\right)$
$$
\begin{array}{l}
=\left[\left(2-\alpha_{1}\right)+\left(2-\alpha_{2}\right)+\cdots+\left(2-\alpha_{n}\right)\right] \cdot\left(\frac{1}{2-\alpha_{1}}+\frac{1}{2-\alpha_{2}}+\cdots+\frac{1}{2-\alpha_{n}}\right) \\
\geqslant\left[\sqrt{2-\alpha_{1}} \cdot \frac{1}{\sqrt{2-\alpha_{1}}}+\sqrt{2-\alpha_{2}} \cdot \frac{1}{\sqrt{2-\alpha_{2}}}+\cdots+\sqrt{2-\alpha_{n}} \cdot \frac{1}{\sqrt{2-\alpha_{n}}}\right]^{2}=n^{2},
\end{array}
$$

Therefore, $y+n \geqslant \frac{2 n^{2}}{2 n-1}, y \geqslant \frac{2 n^{2}}{2 n-1}-n=\frac{n}{2 n-1}$. The equality holds if and only if $\alpha_{1}=\alpha_{2}=\cdots=\alpha_{n}=\frac{1}{n}$, thus $y$ has a minimum value of $\frac{n}{2 n-1}$.