6. Find the function $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$ that satisfies for any three distinct positive real numbers $a, b, c$, the segments of lengths $a, b, c$ can form a triangle if and only if the segments of lengths $f(a), f(b), f(c)$ can form a triangle.
6. If three segments of lengths $u$, $v$, and $w$ can form a triangle, we briefly say “$u$, $v$, $w$ can form a triangle”; otherwise, we say “$u$, $v$, $w$ cannot form a triangle”.
Let the function $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$ satisfy the given conditions.
To derive the conclusion of this problem, we will prove the following five conclusions in sequence.
(1) $f$ is strictly increasing.
Proof by contradiction.
Assume there exist positive real numbers $a < b$ such that $f(a) \geqslant f(b)$.
Thus, $f(x) \leqslant f(b)$ and $f(y) \leqslant f(b)$ can hold for at most one of $x \in I$.
This implies that at most one $x \in I$ satisfies $f(x) \leqslant f(b)$.
Let $I_{1}=\{x \in I \mid f(x)>f(b)\}$.
Clearly, $I_{1}$ is an infinite set.
Assume the set $f\left(I_{1}\right)=\left\{f(x) \mid x \in I_{1}\right\}$ has an upper bound $M$.

Then for any $x, y \in I_{1} (x \neq y)$, by the fact that $f(x)$, $f(y)$, and $f(b)$ cannot form a triangle, we have
$|f(x)-f(y)| \geqslant f(b) > 0$.
Thus, $f\left(I_{1}\right)$ contains at most $\left[\frac{M}{f(b)}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$) distinct elements, and each element has a unique preimage under $f$.
This implies that $I_{1}$ is a finite set, which is a contradiction.
Therefore, the set $f\left(I_{1}\right)$ has no upper bound.
Thus, there exist $x, y \in I_{1}, x \neq y$, such that
$f(x) > f(a)$, and $f(y) > f(x) + f(a)$, but this contradicts the fact that $f(x)$, $f(y)$, and $f(a)$ can form a triangle.
Hence, $f$ is strictly increasing.
(2) If $\left\{x_{n}\right\}$ is a strictly decreasing sequence of positive numbers, and $\lim _{n \rightarrow+\infty} x_{n}=0$, then $\lim _{n \rightarrow+\infty} f\left(x_{n}\right)=0$.

By the conclusion of (1), $\left\{f\left(x_{n}\right)\right\}$ is also a strictly decreasing sequence of positive numbers.
Thus, $\lim _{n \rightarrow+\infty} f\left(x_{n}\right)$ exists, denoted by $A=\lim _{n \rightarrow+\infty} f\left(x_{n}\right) \geqslant 0$.
Assume $A > 0$. Then there exists a positive integer $N$ such that for $n > N$,
$A > f\left(x_{n}\right) > 0$.
By the conclusion of (1), $\left\{f\left(x_{n}\right)\right\}$ is also strictly decreasing, and $f\left(x_{n}\right) > f(y)$.
Thus, $\lim _{n \rightarrow+\infty} f\left(x_{n}\right)$ exists, denoted by $A=\lim _{n \rightarrow+\infty} f\left(x_{n}\right) \geqslant f(y)$.
Consider $x_{n+1}$, $y$, and $x_{n} - y$.
By the condition of $\left\{x_{n}\right\}$, when $n$ is sufficiently large, $0 < x_{n+1} < y < x_{n}$,
Thus, the distinct positive numbers $x_{n+1}$, $y$, and $x_{n} - y$ can form a triangle. Hence,
$$
f\left(x_{n} - y\right) + f(y) > f\left(x_{n+1}\right).
$$

By $\lim _{n \rightarrow+\infty}\left(x_{n} - y\right) = 0$, and the conclusion of (2), we have $\lim _{n \rightarrow+\infty} f\left(x_{n} - y\right) = 0$.
Taking the limit as $n \rightarrow +\infty$ on both sides of (1), we get $f(y) \geqslant \lim _{n \rightarrow+\infty} f\left(x_{n+1}\right) = A$.
Therefore, $\lim _{n \rightarrow+\infty} f\left(x_{n}\right) = A = f(y)$.
(4) For any $x, y > 0$, we have
$f(x + y) = f(x) + f(y)$.
Take a strictly decreasing sequence $\left\{y_{n}\right\}$ such that $\lim _{n \rightarrow+\infty} y_{n} = y$, and all terms of $\left\{y_{n}\right\}$ are less than $x + y$ and not equal to $x$. Thus, $x$, $y_{n}$, and $x + y$ are distinct and form a triangle.
Therefore, $f(x) + f\left(y_{n}\right) > f(x + y)$.
Taking the limit as $n \rightarrow +\infty$ and using the conclusion of (3), we get
$f(x) + f(y) \geqslant f(x + y)$.
Also, $x$, $y_{n}$, and $x + y_{n}$ are distinct and do not form a triangle, so
$f(x) + f\left(y_{n}\right) \leqslant f\left(x + y_{n}\right)$.
Taking the limit as $n \rightarrow +\infty$, we get $f(x) + f(y) \leqslant f(x + y)$.
Therefore, $f(x + y) = f(x) + f(y)$.
(5) Let $f(1) = c > 0$, then $f(x) = c x$.
By the conclusion of (4) and the Cauchy method, for any positive rational number $q$, we have
$$
f(q) = q f(1) = c q.
$$

For any positive real number $x$, take a strictly decreasing sequence of rational numbers $\left\{q_{n}\right\}$ such that $\lim _{n \rightarrow+\infty} q_{n} = x$.
By the conclusion of (2), we have
$f(x) = \lim _{n \rightarrow+\infty} f\left(q_{n}\right) = \lim _{n \rightarrow+\infty} c q_{n} = c x$.
Clearly, $f(x) = c x$ satisfies the conditions.
In summary, the function we seek is
$f(x) = c x$ (where $c$ is any positive real number).
(Xiong Bin provided)