A point $P$ is chosen in an arbitrary triangle. Three lines are drawn through $P$ which are parallel to the sides of the triangle. The lines divide the triangle into three smaller triangles and three parallelograms. Let $f$ be the ratio between the total area of the three smaller triangles and the area of the given triangle. Prove that $f\ge\frac{1}{3}$ and determine those points $P$ for which $f =\frac{1}{3}$ .
1. **Choose a side \( S \) of the triangle and denote the lengths of the sides of the three smaller triangles parallel to \( S \) as \( a, b, \) and \( c \).**

2. **The length of the side of the given triangle parallel to \( S \) is \( a + b + c \).**

3. **The area of a triangle is proportional to the square of the length of its sides. Therefore, the areas of the three smaller triangles are proportional to \( a^2, b^2, \) and \( c^2 \).**

4. **Let \( A \) be the area of the original triangle. The total area of the three smaller triangles is proportional to \( a^2 + b^2 + c^2 \).**

5. **The ratio \( f \) between the total area of the three smaller triangles and the area of the given triangle is given by:**
   \[
   f = \frac{a^2 + b^2 + c^2}{(a + b + c)^2}
   \]

6. **To prove that \( f \geq \frac{1}{3} \), we use the Arithmetic Mean-Quadratic Mean (AM-QM) inequality. The AM-QM inequality states that for any non-negative real numbers \( x_1, x_2, \ldots, x_n \):**
   \[
   \frac{x_1 + x_2 + \cdots + x_n}{n} \leq \sqrt{\frac{x_1^2 + x_2^2 + \cdots + x_n^2}{n}}
   \]

7. **Applying the AM-QM inequality to \( a, b, \) and \( c \), we get:**
   \[
   \frac{a + b + c}{3} \leq \sqrt{\frac{a^2 + b^2 + c^2}{3}}
   \]

8. **Squaring both sides of the inequality, we obtain:**
   \[
   \left( \frac{a + b + c}{3} \right)^2 \leq \frac{a^2 + b^2 + c^2}{3}
   \]

9. **Multiplying both sides by 3, we get:**
   \[
   \frac{(a + b + c)^2}{3} \leq a^2 + b^2 + c^2
   \]

10. **Rearranging the inequality, we have:**
    \[
    \frac{a^2 + b^2 + c^2}{(a + b + c)^2} \geq \frac{1}{3}
    \]

11. **Thus, we have shown that \( f \geq \frac{1}{3} \).**

12. **Equality holds in the AM-QM inequality if and only if \( a = b = c \). In the context of the problem, this means that the point \( P \) must be such that the three smaller triangles are congruent. This occurs when \( P \) is the centroid of the triangle.**

\(\blacksquare\)

The final answer is \( \boxed{ f \geq \frac{1}{3} } \) and \( f = \frac{1}{3} \) when \( P \) is the centroid.