Let $x \leq y \leq z \leq t$ be real numbers such that $xy + xz + xt + yz + yt + zt = 1.$ 

[b]a)[/b] Prove that $xt < \frac 13,$

b) Find the least constant $C$ for which inequality $xt < C$ holds for all possible values $x$ and $t.$
### Part (a)
1. **Assume the contrary**: Suppose \( xt \geq \frac{1}{3} \). We need to show that this leads to a contradiction.
2. **Consider the sum**: Given \( xy + xz + xt + yz + yt + zt = 1 \), and \( x \leq y \leq z \leq t \).
3. **Analyze the positivity**: Since \( x \leq y \leq z \leq t \), if \( x \) and \( t \) are both positive, then \( y \) and \( z \) are also positive.
4. **Sum of terms**: The sum \( xy + xz + xt + yz + yt + zt \) includes the terms \( xt + yt + zt \). Since \( x \leq y \leq z \leq t \), we have:
   \[
   xt + yt + zt \geq 3xt
   \]
5. **Contradiction**: If \( xt \geq \frac{1}{3} \), then:
   \[
   3xt \geq 3 \cdot \frac{1}{3} = 1
   \]
   This implies:
   \[
   xt + yt + zt \geq 1
   \]
   However, this would mean that the sum \( xy + xz + xt + yz + yt + zt \) is greater than or equal to 1, which contradicts the given condition \( xy + xz + xt + yz + yt + zt = 1 \).

6. **Negative case**: If \( x \) and \( t \) are both negative, consider the transformation \( P(-t, -z, -y, -x) \). The same argument applies because the sum of products remains the same, and the inequality \( xt < \frac{1}{3} \) holds.

Thus, we have shown that \( xt < \frac{1}{3} \).

### Part (b)
1. **Construct a specific example**: Let \( x = y = z = r \) and \( t = \frac{1}{3r} - r \) with \( 0 < r < \frac{1}{3} \).
2. **Verify the inequality**: We need to check if \( x \leq y \leq z \leq t \) holds. Since \( x = y = z = r \) and \( t = \frac{1}{3r} - r \), we have:
   \[
   r \leq \frac{1}{3r} - r
   \]
   Simplifying, we get:
   \[
   r + r \leq \frac{1}{3r}
   \]
   \[
   2r \leq \frac{1}{3r}
   \]
   \[
   6r^2 \leq 1
   \]
   \[
   r^2 \leq \frac{1}{6}
   \]
   Since \( 0 < r < \frac{1}{3} \), this inequality holds.
3. **Calculate \( xt \)**: We have:
   \[
   xt = r \left( \frac{1}{3r} - r \right) = \frac{1}{3} - r^2
   \]
4. **Limit as \( r \to 0 \)**: As \( r \) approaches 0, \( xt \) approaches:
   \[
   \frac{1}{3} - r^2 \to \frac{1}{3}
   \]
   Therefore, the least constant \( C \) for which \( xt < C \) holds for all possible values of \( x \) and \( t \) is \( \frac{1}{3} \).

The final answer is \( \boxed{\frac{1}{3}} \)