Problem: Determine all real constants $t$ such that whenever $a$, $b$ and $c$ are the lengths of sides of a triangle, then so are $a^2+bct$, $b^2+cat$, $c^2+abt$.

Solution: \paragraph{Answer.}
We will show the answer is exactly \[ 2/3 \le t \le 2. \]

\paragraph{Proof.}
Apply the substitution $a=y+z$, $b=z+x$, $c=x+y$.
We need the inequality
\begin{align*}
  a^2 + bct & < b^2 + ca t + c^2 + ab t \\
  (y+z)^2 + (x+y)(x+z) t &< (x+y)^2+(x+z)^2 + (y+z)(2x+y+z) t \\
  \iff [x^2-(xy+xz+yz+y^2+z^2)]t &< 2x^2+2xy+2xz-2yz \\
\end{align*}
to be true for all $(x,y,z) \in \RR_{>0}^3$
(checking only one inequality is enough by symmetry).
Writing this is a quadratic in $x$, we want
\[ Q(x) \coloneq (2-t)x^2 + (2+t)(y+z)x + ((y^2+yz+z^2)t - 2yz) > 0. \]

\begin{claim*}
  Except when $t=2$,
  the quadratic $Q$ always has two real roots.
\end{claim*}
\begin{proof}
The discriminant is
\begin{align*}
  D &= [(2+t)(y+z)]^2 - 4(2-t)\left( (y^2+yz+z^2)t - 2yz \right) \\
  &= \left( (2+t)^2 - 4(2-t)t \right)(y^2+z^2)
  + \left( 2(2+t)^2-4(2-t)(t-2) \right) yz \\
  &= \left( 5t^2-4t+4 \right)(y^2+z^2) + \left( 6t^2-8t+24  \right) yz > 0
\end{align*}
which means $Q$ always has two real roots
(aside for the single exceptional case $t=2$,
in which $Q$ is not a quadratic).
\end{proof}

Now, we make a few closing observations.
\begin{itemize}
  \ii It is clear we need $t \le 2$,
  since a negative leading coefficient will
  cause the inequality to fail for $x \gg y,z$.
  \ii The case $t=2$ obviously has $Q > 0$ always.
  \ii For $t < 2$, as $Q$ always has two real roots,
  the assertion is true if and only if all coefficients
  of $Q$ are nonnegative.
  \ii When $t \ge \frac23$,
  we have $(2+t)(y+z) > 0$ obviously, and from
  \[ \frac{y^2+yz+z^2}{3} \ge yz \]
  the constant coefficient is nonnegative as well.
  Thus when $\boxed{\frac23 \le t \le 2}$
  we indeed have $Q(x) > 0$ for $x,y,z > 0$.
  \ii If $t < \frac23$,
  then by letting $y=z$ fails the condition.
\end{itemize}