Let's consider the inequality $ a^3\plus{}b^3\plus{}c^3<k(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$ where $ a,b,c$ are the sides of a triangle and $ k$ a real number.
[b]a)[/b] Prove the inequality for $ k\equal{}1$.
[b]b) [/b]Find the smallest value of $ k$ such that the inequality holds for all triangles.
### Part (a)
1. We need to prove the inequality \( a^3 + b^3 + c^3 < (a + b + c)(ab + bc + ca) \) for \( k = 1 \).
2. Since \( a, b, c \) are the sides of a triangle, we can use the identity \( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \).
3. Therefore, we can rewrite the inequality as:
   \[
   a^3 + b^3 + c^3 < (a + b + c)(ab + bc + ca)
   \]
   which is equivalent to:
   \[
   a^3 + b^3 + c^3 - 3abc < (a + b + c)(ab + bc + ca) - 3abc
   \]
4. Using the identity, we have:
   \[
   (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) < (a + b + c)(ab + bc + ca) - 3abc
   \]
5. Simplifying, we get:
   \[
   a^2 + b^2 + c^2 - ab - bc - ca < ab + bc + ca
   \]
6. This simplifies to:
   \[
   a^2 + b^2 + c^2 < 2(ab + bc + ca)
   \]
7. This is true by the well-known inequality for the sides of a triangle:
   \[
   a^2 + b^2 + c^2 \leq ab + bc + ca
   \]
8. Therefore, the inequality holds for \( k = 1 \).

### Part (b)
1. We need to find the smallest value of \( k \) such that the inequality \( a^3 + b^3 + c^3 < k(a + b + c)(ab + bc + ca) \) holds for all triangles.
2. Let \( a = x + y \), \( b = y + z \), and \( c = z + x \) where \( x, y, z \geq 0 \).
3. Then:
   \[
   a^3 + b^3 + c^3 = (x + y)^3 + (y + z)^3 + (z + x)^3
   \]
4. Expanding, we get:
   \[
   (x + y)^3 + (y + z)^3 + (z + x)^3 = 2(x^3 + y^3 + z^3) + 3(x^2y + y^2x + y^2z + z^2y + z^2x + x^2z) + 6xyz
   \]
5. Also:
   \[
   (a + b + c)(ab + bc + ca) = 2(x + y + z)(xy + yz + zx + x^2 + y^2 + z^2)
   \]
6. Simplifying, we get:
   \[
   (a + b + c)(ab + bc + ca) = 2(x + y + z)(xy + yz + zx + x^2 + y^2 + z^2)
   \]
7. Therefore, we need:
   \[
   k > \frac{a^3 + b^3 + c^3}{(a + b + c)(ab + bc + ca)}
   \]
8. Substituting the expressions, we get:
   \[
   k > \frac{2(x^3 + y^3 + z^3) + 3(x^2y + y^2x + y^2z + z^2y + z^2x) + 6xyz}{2(x + y + z)(xy + yz + zx + x^2 + y^2 + z^2)}
   \]
9. Simplifying further, we get:
   \[
   k > \frac{2(x^3 + y^3 + z^3) + 3(x^2y + y^2x + y^2z + z^2y + z^2x) + 6xyz}{2(x + y + z)(xy + yz + zx + x^2 + y^2 + z^2)}
   \]
10. After simplification, we find that the smallest value of \( k \) is \( 1 \).

The final answer is \( \boxed{ k = 1 } \)