Let a, b, c be the sides of a triangle. If t | vedclass

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(C) Let $a, b, c$ be the sides of a triangle. Since $a^2+b^2 \geq 2ab$,$b^2+c^2 \geq 2bc$,and $c^2+a^2 \geq 2ac$,adding these inequalities gives $2(a^

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$\{x \in \mathbb{R} \mid 1 \leq x < 2\}$

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(C) Let $a, b, c$ be the sides of a triangle. Since $a^2+b^2 \geq 2ab$,$b^2+c^2 \geq 2bc$,and $c^2+a^2 \geq 2ac$,adding these inequalities gives $2(a^2+b^2+c^2) \geq 2(ab+bc+ca)$,which implies $\frac{a^2+b^2+c^2}{ab+bc+ca} \geq 1$. Thus,$t \geq 1$. For a triangle,the triangle inequality states $a+b > c$,$b+c > a$,and $c+a > b$. Squaring $a+b > c$ gives $a^2+b^2+2ab > c^2$,or $a^2+b^2-c^2 > -2ab$. This does not directly bound $t$. However,consider $(a-b)^2 + (b-c)^2 + (c-a)^2 > 0$ for distinct sides,which leads to $a^2+b^2+c^2 > ab+bc+ca$,so $t > 1$. Using the triangle inequality $a Similarly,$b^2 Adding these,$a^2+b^2+c^2 Combining these,the set of values is $1 \leq t Therefore,the set is $\{x \in \mathbb{R} \mid 1 \leq x < 2\}$.

Let $a, b, c$ be the sides of a triangle. If $t$ denotes the expression $\frac{a^2+b^2+c^2}{ab+bc+ca}$,the set of all possible values of $t$ is

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