Determine all real constants $t$ such that whenever $a$, $b$ and $c$ are the lengths of sides of a triangle, then so are $a^2+bct$, $b^2+cat$, $c^2+abt$.
1. **Identify the problem and given conditions:**
   We need to determine all real constants \( t \) such that if \( a, b, \) and \( c \) are the lengths of the sides of a triangle, then \( a^2 + bct, b^2 + cat, \) and \( c^2 + abt \) are also the lengths of the sides of a triangle.

2. **Necessary conditions:**
   - Consider the triangle inequality conditions for the new sides:
     \[
     a^2 + bct + b^2 + cat > c^2 + abt
     \]
     \[
     a^2 + bct + c^2 + abt > b^2 + cat
     \]
     \[
     b^2 + cat + c^2 + abt > a^2 + bct
     \]

3. **Verification for specific values:**
   - **Case 1:** Let \( a = b = 1 \) and \( c = 2 - \varepsilon \):
     \[
     (a^2 + bct, b^2 + cat, c^2 + abt) = (1 + (2 - \varepsilon)t, 1 + (2 - \varepsilon)t, (2 - \varepsilon)^2 + t)
     \]
     We require:
     \[
     2(1 + 2t) > 4 - \varepsilon + t \implies 3t > 2 - \varepsilon \implies t \geq \frac{2}{3}
     \]

   - **Case 2:** Let \( a = \varepsilon \) and \( b = c = 1 \):
     \[
     (a^2 + bct, b^2 + cat, c^2 + abt) = (\varepsilon^2 + t, 1 + \varepsilon t, 1 + \varepsilon t)
     \]
     We require:
     \[
     2(1 + \varepsilon t) > \varepsilon + t \implies t \leq 2
     \]

4. **Verification that \( t \in \left[\frac{2}{3}, 2\right] \) works:**
   - Without loss of generality, assume \( a \leq b \leq c \). The primary condition to check is \( a + b > c \).
   - Take the derivative of the expressions with respect to \( c \):
     \[
     (a^2 + bct)' = bt
     \]
     \[
     (b^2 + cat)' = at
     \]
     \[
     (c^2 + abt)' = 2c
     \]
   - For the inequalities \( c^2 + abt + a^2 + bct > b^2 + cat \) and \( c^2 + abt + b^2 + act > a^2 + bct \), we have:
     \[
     (LHS)' = 2c \geq 2b \geq (RHS)'
     \]
     Thus, increasing \( c \) weakens the inequality, so we only need to check \( c = b \).

5. **Check \( c = b \):**
   - The inequalities become:
     \[
     a^2 + b^2 + abt + b^2t > b^2 + abt
     \]
     \[
     2b^2 + 2abt > a^2 + b^2t
     \]
     The first is obvious, and the second holds as \( 2b^2 \geq b^2t \) and \( 2abt \geq 2a^2t > a^2 \).

6. **Check \( c = a + b - \varepsilon \):**
   - The inequality becomes:
     \[
     a^2 + abt + b^2t + b^2 + a^2t + abt - \varepsilon(a + b)t > a^2 + 2ab + b^2 - 2\varepsilon(a + b) + \varepsilon^2 + abt
     \]
     Simplifying:
     \[
     t(a^2 + ab + b^2) + \varepsilon(2 - t)(a + b) > 2ab + \varepsilon^2
     \]
     If \( t = 2 \), then:
     \[
     2(a^2 + ab + b^2) \geq 6ab > 2ab + \varepsilon^2
     \]
     Otherwise, for small enough \( \varepsilon \):
     \[
     \varepsilon(2 - t)(a + b) > \varepsilon^2
     \]
     It suffices to show that \( t(a^2 + ab + b^2) \geq 2ab \), which is true by AM-GM and \( t \geq \frac{2}{3} \).

Thus, the advertised values of \( t \) indeed work, so we're done. \(\blacksquare\)

The final answer is \( \boxed{ t \in \left[\frac{2}{3}, 2\right] } \)