6. Find the smallest real number $r$ such that a rod of length 1 can be broken into four parts, each of length at most $r$, so that no three of these parts can form a triangle.
SOLUTION. The only criterion for whether we can construct a triangle with sides of given lengths is the triangle inequality. To construct a triangle, it is sufficient for the largest length to be
smaller than the sum of the remaining two. If this inequality holds, the other triangle inequalities also hold, and the triangle can be constructed. If it does not hold, the triangle cannot be constructed because the triangle inequality is not satisfied. A triangle cannot be constructed from three sides of given lengths precisely when the largest length is at least as large as the sum of the remaining two.

Assume that $r$ is the smallest number we are looking for. This means that we can break the stick into four parts, each of length at most $r$, such that no triangle can be formed from any three of these parts. The longest part must have length $r$, otherwise $r$ would not be the smallest number with the required property. Let the lengths of the other parts be $x, y, z, r$ such that $x \leq y \leq z \leq r$ and $x+y+z+r=1$, and assume that no triangle can be formed from any three such parts, which we now express with the appropriate inequalities.

A triangle cannot be constructed from the lengths $(x, y, z)$ precisely when $x+y \leq z$. Since $z \leq r$, it follows that the inequality $x+y \leq r$ also holds for the lengths $(x, y, r)$, so no triangle can be formed from these three lengths either. Similarly, for the lengths $(y, z, r)$, a triangle cannot be formed precisely when $y+z \leq r$, and since $x \leq y$, it follows that the inequality $x+z \leq r$ also holds for the lengths $(x, z, r)$.

Therefore, no triangle can be formed from any three parts precisely when, in addition to the equality $x+y+z+r=1$, both conditions

$$
x+y \leq z \quad \text{and} \quad y+z \leq r
$$

are satisfied.

If we substitute $x+y=1-z-r$ into the first inequality (1), we get

$$
\begin{aligned}
1-z-r & \leq z \\
1-r & \leq 2 z .
\end{aligned}
$$

Since $y \geq x$, the second inequality from (1) has the following consequence:

$$
r \geq y+z=\frac{y+y+z}{2}+\frac{z}{2} \geq \frac{x+y+z}{2}+\frac{z}{2}=\frac{1-r}{2}+\frac{z}{2} .
$$

In the obtained inequality, we further estimate $z$ using inequality (2), so we get

$$
r \geq \frac{1-r}{2}+\frac{z}{2} \geq \frac{1-r}{2}+\frac{1-r}{4} \geq \frac{3}{4}(1-r),
$$

from which, by comparing the right and left sides, the required estimate of the value of $r$ follows:

$$
4 r \geq 3(1-r) \quad \text{or} \quad r \geq \frac{3}{7} .
$$

It remains to show that there is a way to break the stick of length 1 into four parts, each of length at most $3 / 7$, such that no triangle can be formed from any three of these parts - for example, the lengths $(1 / 7, 1 / 7, 2 / 7, 3 / 7)$ satisfy this condition.