Let $a, b, c$ be the sides of a triangle.
$\therefore \quad a^2+b^2 \geq 2 a b \quad[\because AM \geq GM ]$
Similarly, $b^2+c^2 \geq 2 b c$
$c^2+a^2 \geq 2 a c$
$\Rightarrow 2\left(a^2+b^2+c^2\right) \geq 2(a b+b c+c a)$
$\Rightarrow \quad \frac{a^2+b^2+c^2}{a b+b c+a c} \geq 1$
$t \geq 1 \quad\left[\because \frac{a^2+b^2+c^2}{a b+b c+a c}=t\right]$
$\because a, b, c$ be the side of a triangle.
$\begin{aligned} a+b>c \\ & \quad|a|>|c-b| \\ a \end{aligned}$
$\Rightarrow \quad a^2 > (c-b)^2$
$\Rightarrow \quad a^2 > c^2+b^2-2 b c$
$\Rightarrow \quad b^2+c^2-a^2<2 b c$
Similarly, $a^2+b^2-c^2 < 2 a b$
and $\quad c^2+a^2-b^2 < 2 a c$
On adding Eqs.$(i), (ii)$ and $(iii)$, we get
$a^2+b^2+c^2 < 2(a b+b c+c a)$
$\Rightarrow \quad \frac{a^2+b^2+c^2}{a b+b c+c a} < 2$
$\therefore \quad t < 2$
$\therefore \quad\{x \in R \mid 1 \leq x < 2\}$
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