If we allow degenerate triangles for a second then taking $b=c$ and $a=2b$ gives that $a^{2}+bct=b^{2}\left(4+t\right),b^{2}+cat=c^{2}+abt=b^{2}\left(1+2t\right)$. Thus we need $4+t\leq2\left(1+2t\right)\Rightarrow t\geq\frac{2}{3}$. Also taking $a=b,c=0$ gives that $a^{2}+bct=b^{2}+cat=b^{2},c^{2}+abt=tb^{2}$. Thus we need $t\leq2$. i.e. we need $\frac{2}{3}\leq t\leq2$. N.B. I know that we may not be allowed degenerate triangles but the argument above can be made rigorous by saying that if $t<\frac{2}{3}$ then by picking $b=c$ and $a$ just less than $2b$ then we get that $a^2+bct,b^2+cat,c^2+abt$ do not form the sides of a triangles and similarly for $t>2$. Now we must show that $\frac{2}{3}\leq t\leq2$ suffices. If $a,b,c$ are the side lengths of a triangle then $\exists x,y,z\in\mathbb{R}^{+}$ such that $a=y+z,b=z+x,c=x+y$. Then: $a^{2}+bct=y^{2}+z^{2}+2yz+\left(x^{2}+xy+yz+zx\right)t,$ $b^{2}+cat=z^{2}+x^{2}+2zx+\left(y^{2}+yz+zx+xy\right)t,$ $c^{2}+abt=x^{2}+y^{2}+2xy+\left(z^{2}+zx+xy+yz\right)t$ we need to show that $b^{2}+cat+c^{2}+abt>a^{2}+bct$ as then by symmetry we are done. But this is equivalent to: $2x^{2}+2zx+2xy+\left(y^{2}+z^{2}+yz+zx+xy\right)t>2yz+x^{2}t$ Now as $t\leq2:$ $2x^{2}\geq x^{2}t$ and as $t\geq\frac{2}{3}:$ $\left(y^{2}+z^{2}+yz+zx+xy\right)t>\left(y^{2}+z^{2}+yz\right)t\geq3yzt\geq2yz$ As required. $\square$

You could just consider discriminant of two quadratic equations, then you could get $ 2\ge t \ge \frac{2}{3} $

Rather boring problem IMO. A brute-force solution: it's necessary and sufficient (by symmetry) that $a^2+tbc < \frac12 \textstyle \sum a^2+tbc$ for all triangles $ABC$. Then just put $a=y+z$, et cetera, and expand.

$a^2+bct<b^2+act+c^2+abt$ and $\implies b^2+cat<p$ $\implies c^2+abt<k$ , $k, p\in\mathbb{Z}$ $a=y+z,b=z+x,c=x+y$ $(y+z)^2+(z+x)(x+y)t<(z+x)^2+act+c^2+abt$ And this inequality can be solved by quadratic equation $D>0$ ,$b^2–4ac\ge 0$ or by simply quadratic in $x$ . $\frac{2}{3}\le t \le 2$.

Solution with stroller. We have the bounds (and symmetric versions thereof) \[a+b>c,a^2+b^2+(a+b)ct > c^2+abt.\]We claim that $t$ works iff $t\in [2/3,2]$. We have $t \ge 2/3$ by taking $a=b=1+\epsilon, c=2$ and small $\epsilon$, since we get \[2+4\epsilon + 2\epsilon^2 + (4+4\epsilon)t > 4+(1+2\epsilon+\epsilon^2)t \implies 2-4\epsilon-2\epsilon^2 < (3-2\epsilon - \epsilon^2)t \implies \]\[t > \frac{2}{3} - \frac{4}{3} \cdot \frac{2\epsilon + 1}{3-2\epsilon - \epsilon^2}.\]We have $t \le 2$ by taking $a=b=1,c=\epsilon$, since we get \[2 + 2\epsilon t > \epsilon^2 + t \implies 2-t > \epsilon (\epsilon - 2t).\]This suffices because $\lim_{\epsilon \rightarrow 0} \epsilon (\epsilon - 2t)= 0$. Now, we show that these upper and lower bounds suffice. By the linearity of $a^2+bct,b^2+cat,c^2+abt$ in $t$, it suffices to check $t=2$ and $t=2/3$. To check $t=2$, observe that we have \[(a-b)^2\ge 0 > c(c-2a-2b).\]Now, we check $t=2/3$. We observe that \[a^2 + c^2 - b^2 > \frac{2}{3}(ac -bc - ab)\]\[\iff a^2 + c^2 - b^2 - \frac{2}{3}ac + \frac{2}{3}bc + \frac{2}{3}ab > 0\]\[\iff \frac{1}{3}(a-c)^2 + \frac{2}{3}(a^2 + c^2) - b^2 + \frac{2}{3}bc + \frac{2}{3}ab > 0.\]It suffices to show that \[2(a^2 + c^2) - 3b^2 + 2b(a+c)> 0.\]Indeed, let $s = a+c$ and note that \[2(a^2 + c^2) - 3b^2 + 2b(a+c) \ge s^2 + 2bs - 3b^2 > b^2 + 2b \cdot b - 3b^2 = 0.\]

Plugging in $(a,b,c)=(1,1,0)$ (or something very close to this) gives $1,1,t$ are a triangle; hence $t<2$. Plugging in $(a,b,c)=(1,1,2)$ gives $1+2t,1+2t,4+t$ are a triangle; hence $2+4t > 4+t$, i.e. $t > 2/3$. We claim any $2/3 < t < 2$ works. It suffices to find all $t$ such that \[ 2(a^2+bct) < (a^2+b^2+c^2)+(ab+bc+ca)t. \]Let $a=y+z,b=z+x,c=x+y$, which encodes the $a,b,c$ triangle condition. Then \begin{align*} &\qquad 2(y^2+2yz+z^2+(x^2+xy+xz+yz)t) \\ &< 2(x^2+y^2+z^2+xy+yz+zx) + t(x^2+y^2+z^2+3xy+3yz+3zx) \\ \iff &\qquad 2tx^2+2y^2+2z^2 +2txy+2txz+(2t+4)yz \\ &< (2+t)x^2+(2+t)y^2+(2+t)z^2+(3t+2)xy+(3t+2)yz+(3t+2)zx \\ \iff &0< (2-t)x^2+ty^2+tz^2+ (t+2)xy+ (t-2)yz+(t+2)zx. \end{align*}Since the above inequality is linear in $t$, it suffices to check that it holds for $t=2/3$ and $t=2$; then anything in between holds. For $t=2$, we have \[ 0<2y^2+2z^2 + 4xy + 4zx,\]which is true since $x,y,z$ are positive. For $t=2/3$, the inequality becomes \begin{align*} 0&<4 x^2 + 2 y^2 + 2 z^2 + 8 xy - 4 yz + 8 zx \\ &= 4x^2 + 2(y-z)^2 + 8xy+8zx, \end{align*}which is true. This completes the proof.

The answer is all $t \in \left[ \frac{2}{3} , 2 \right]$. WLOG $a \ge b \ge c$. We want \begin{align*} a^2 + bct < b^2 + c^2 + at(b+c) \qquad \qquad (1) \\ b^2 + cat < a^2 + c^2 + bt (c+a) \qquad \qquad (2) \\ c^2 + abt < a^2 + b^2 + ct(a+b) \qquad \qquad (3) \end{align*}First we show $\frac{2}{3} \le t \le 2$. Observe: In $(1)$, fixing $b=c=k$ and choosing $a = 2k - \epsilon$ for a sufficiently small $\epsilon > 0$ forces $t \ge \frac{2}{3}$. In $(3)$, fixing $a=b=k$ and choosing $c = \epsilon$ for a sufficiently small $\epsilon > 0$ forces $t \le 2$. Now assume $\frac{2}{3} \le t \le 2$. We will show all $(1),(2),(3)$ are true. $(2)$ is easily true. Now we prove $(1)$. Let $b+c = 2m$. Then, $$b^2 + c^2 \ge 2m^2 ~~,~~ t(ab + ac - bc) \ge t ( 2am - m^2) \ge \frac{2}{3} \cdot (2am - m^2) $$So it suffices to show $$a^2 < 2m^2 + \frac{4am}{3} - \frac{2m^2}{3} \iff (2m-a) \left( \frac{2m + 3a}{3} \right) > 0$$which is true as $2m - a > 0$ by triangle inequality. Now we prove $(3)$. Note $$a^2 + b^2 + bct + cat \ge 2ab + 2tc^2$$So it suffices to show $$2ab + 2c^2 \cdot t > c^2 + abt \iff ab(2-t) + c^2 (2t-1) > 0$$Writing $ab(2-t) \ge c^2(2t-1)$, we obtain it suffices to show $$0 < c^2(2-t) + c^2(2t-1) = c^2(t+1)$$This proves $(3)$, completing the proof.$\blacksquare$

By symmetry we just need $a^2+b^2+bct+cat>c^2+abt$ for $t$ to be satisfied. Now apply the Ravi substitution $a=x+y$, $b=y+z$, $c=z+x$ for $x,y,z\in\mathbb R^+$. We have, after expanding: $$\Leftrightarrow2xy+2y^2+2yz+(xy+yz+zx+x^2-y^2+z^2)t>2xz.$$ We first claim that $\frac23\le t\le2$. Setting $x=z=1$ and $y=\varepsilon$ for $0<\varepsilon<3$ is: $$t>\frac{2-4\varepsilon-2\varepsilon^2}{3+2\varepsilon-\varepsilon^2}$$and since $\lim_{\varepsilon\to0^+}\frac{2-4\varepsilon-2\varepsilon^2}{3+2\varepsilon-\varepsilon^2}=\frac23$, $t\ge\frac23$. Setting $x=z=\varepsilon$ and $y=1$ for $0<\varepsilon<\frac13$ gives: $$t<\frac{2\varepsilon^2-4\varepsilon-2}{3\varepsilon^2+2\varepsilon-1}$$and since $\lim_{\varepsilon\to0^+}\frac{2\varepsilon^2-4\varepsilon-2}{3\varepsilon^2+2\varepsilon-1}=2$, $t\le2$. If $\frac23\le t\le2$ then we claim that: $$2xy+2y^2+2yz+(xy+yz+zx+x^2-y^2+z^2)t>2xz.$$Since this is a linear inequality wrt $t$, it suffices to show this for $t=\frac23$ and $t=2$. If $t=\frac23$ then: $$2xy+2y^2+2yz+(xy+yz+zx+x^2-y^2+z^2)t-2xz=\frac43y^2+\frac83(xy+yz)+\frac23(x-z)^2>0$$and if $t=2$: $$2xy+2y^2+2yz+(xy+yz+zx+x^2-y^2+z^2)t-2xz=4(xy+yz)+(x+z)^2+(x-z)^2>0.$$

The answer is $t \in [2/3,2]$. To verify that this is necessary, consider the following two triangles: If $a=b=1$ and $c=2-\varepsilon$, we have $(a^2+bct,b^2+cat,c^2+abt)=(1+(2-\varepsilon)t,1+(2-\varepsilon)t,(2-\varepsilon)^2+t)$, so we require $2(1+2t)>4-\varepsilon+t \implies 3t>2-\varepsilon$, or $t \geq 2/3$. If $a=\varepsilon$ and $b=c=1$, we have $(a^2+bct,b^2+cat,c^2+abt)=(\varepsilon^2+t,1+\varepsilon t,1+\varepsilon t)$, so we require $2(1+\varepsilon t)>\varepsilon+t \implies t \leq 2$. Now we verify that $t \in [2/3,2]$ works. WLOG let $a \leq b \leq c$, so the only condition we care about is $a+b>c$. Now we take the derivative of the three expressions of interest with respect to $c$: \begin{align*} (a^2+bct)'&=bt\\ (b^2+cat)'&=at\\ (c^2+abt)'&=2c. \end{align*}For the inequalities $c^2+abt+a^2+bct>b^2+cat$ and $c^2+abt+b^2+act>a^2+bct$, we always have $(LHS)'>2c\geq 2b\geq (RHS)'$, so increasing $c$ only makes the inequality weaker and we thus only need to check $c=b$. In this case, the inequalities become $a^2+b^2+abt+b^2t>b^2+abt$ and $2b^2+2abt>a^2+b^2t$. The first is obvious, and the second holds as $2b^2 \geq b^2t$ and $2abt\geq 2a^2t>a^2$. For the inequality $a^2+bct+b^2+cat>c^2+abt$, if we have $t(a+b)\geq 2c$ then decreasing $c$ only makes the inequality stronger, while if we have $t(a+b)<2c$ then increasing $c$ only makes the inequality stronger. Thus we only have to check $c=b$ and $c=a+b-\varepsilon$ where $\varepsilon$ is sufficiently small. For $c=b$, the inequality becomes $$a^2+c^2t+b^2+abt>b^2+abt,$$which is obvious. For $c=a+b-\varepsilon$, the inequality becomes \begin{align*} 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 & &\iff\\ t(a^2+ab+b^2)+\varepsilon(2-t)(a+b)&> 2ab+\varepsilon^2 \end{align*}If $t=2$, then we have $2(a^2+ab+b^2)\geq 6ab>2ab+\varepsilon^2$. Otherwise, for small enough $\varepsilon$, we have $\varepsilon(2-t)(a+b)>\varepsilon^2$, in which case it suffices to show that $t(a^2+ab+b^2)\geq 2ab$, but this is true by AM-GM and $t \geq 2/3$. Thus, the advertised values of $t$ indeed work, so we're done. $\blacksquare$

Let $a=x+y$, $b=y+z$, and $c=z+x$ for positive $x,y,z$. Then $a^2+bct=x^2+y^2+tz^2+(2+t)xy+tyz+tzx$, $b^2+cat=tx^2+y^2+z^2+txy+(2+t)yz+tzx$, and $c^2+abt=x^2+ty^2+z^2+txy+tyz+(2+t)zx$. Then, we must have that $q+r=a^2+bct=x^2+y^2+tz^2+(2+t)xy+tyz+tzx$, $r+s=b^2+cat=tx^2+y^2+z^2+txy+(2+t)yz+tzx$, and $s+q=c^2+abt=x^2+ty^2+z^2+txy+tyz+(2+t)zx$ for positive $q,r,s$. We solve for $s=\frac{-(q+r)+(r+s)+(s+q)}{2}=\frac{t}{2}(x^2+y^2+z(x+y)+xy-z^2)+z^2-xy+(x+y)z$. If we fix $p=xy$, then the minimum value of $s$ is $(\frac{3t}2-1)p+(t+2)z\sqrt{p}+(1-\frac{t}{2})z^2$ which is achieved when $x=y=\sqrt{p}$ by AM-GM. If $t<\frac23$, then $s$ could be negative for large enough $p$ and if $t>2$, then $s$ could be negative for large enough $z$. Therefore, $t\in [\frac23, 2]$, which clearly works because $p,z>0$.

I originally thought the solution set was $[1, 2]$, but it turned out to be $\left[\frac23, 2\right]$. Let $a+b>c$ so we want $a^2+bct +b^2+act > c^2+abt$, which we can transform to \[(a-b)^2+ab(2-t) +t\cdot c(a+b)>c^2,\]so $t\in [1,2]$ clearly works. Testing $t=1-\varepsilon$, we do some algebraic manipulations to get \[\varepsilon < \frac{\frac14 + 3\varepsilon_2 + \varepsilon_2^2}{\frac34 + \varepsilon_2 - \varepsilon^2_2},\]which gives $\varepsilon \leq \frac{2}{3}$, meaning $t\geq \frac{2}{3}$. Instead, using $t=2+\varepsilon$, doing some algebraic manipulation gets us \[\varepsilon < \frac{4x-1}{x^2-2x}\]for any large $x$, giving $\varepsilon\leq 0$, so $t\leq 2$. The key insight for both is that we can just let $a=b$ to remove that $(a-b)^2$ terms from the LHS, and then use $\frac{a}{c}>\frac12$ for the lower bound, letting $\frac{a}{c}=\frac12 + \varepsilon_2$, and then setting $\frac{a}{c}=x$ to be arbitrarily large to get the upper bound.