Determine all real constants $t$ such that whenever $a, b, c$ are the lengths of the sides of a triangle, then so are $a^{2}+b c t, b^{2}+c a t, c^{2}+a b t$.

Proposed by S. Khan, UNK
The answer is the interval $[2 / 3,2]$.
If $t \geq 2$, then take a triangle with sides $b=c=1$ and $a=\epsilon$. Then $b^{2}+c a t+c^{2}+a b t-a^{2}-b c t=2-t+\epsilon(2 t-\epsilon) \leq 0$ for small positive $\epsilon$; for instance, for any $0 < \epsilon < 1$. Then using $(b+c)^{2} \geq 4 b c$ we obtain

$$
\begin{aligned}
b^{2}+c a t+c^{2}+a b t-a^{2}-b c t & =(b+c)^{2}+a t(b+c)-(2+t) b c-a^{2} \\
& \geq(b+c)^{2}+a t(b+c)-\frac{1}{4}(2+t)(b+c)^{2}-a^{2} \\
& \geq \frac{1}{4}(2-t)(b+c)^{2}+a t(b+c)-a^{2}
\end{aligned}
$$

As $2-t \geq 0$ and $t>0$, this last expression is an increasing function of $b+c$, and hence using $b+c>a$ we obtain

$$
b^{2}+c a t+c^{2}+a b t-a^{2}-b c t>\frac{1}{4}(2-t) a^{2}+t a^{2}-a^{2}=\frac{3}{4}\left(t-\frac{2}{3}\right) a^{2} \geq 0
$$

as $t \geq 2 / 3$. The other two inequalities follow by symmetry.