1. Find all real constants $t$ such that if $a, b, c$ are the side lengths of some triangle, then $a^{2}+b c t, b^{2}+c a t, c^{2}+a b t$ are also the side lengths of some triangle.
1. $t \in\left[\frac{2}{3}, 2\right]$.

If $t > 2$, consider a triangle with side lengths $b=c=1, a=\varepsilon$ for some positive $\varepsilon$ (such as $0 < \varepsilon < 2$). In this case, $a < b+c$.

Combining the inequality $(b+c)^{2} \geqslant 4 b c$, we get
$$
\begin{array}{l}
b^{2}+c a t+c^{2}+a b t-a^{2}-b c t \\
=(b+c)^{2}+a t(b+c)-(2+t) b c-a^{2} \\
\geqslant(b+c)^{2}+a t(b+c)-\frac{1}{4}(2+t)(b+c)^{2}-a^{2} \\
=\frac{1}{4}(2-t)(b+c)^{2}+a t(b+c)-a^{2} .
\end{array}
$$

Given $2-t \geqslant 0, t>0$, the rightmost expression is a monotonically increasing function of $b+c$. Combining $b+c>a$ and $t \geqslant \frac{2}{3}$, we have
$$
\begin{array}{l}
b^{2}+c a t+c^{2}+a b t-a^{2}-b c t \\
>\frac{1}{4}(2-t) a^{2}+t a^{2}-a^{2}=\frac{3}{4}\left(t-\frac{2}{3}\right) a^{2} \geqslant 0 .
\end{array}
$$

By symmetry, the other two inequalities can be derived.