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Question

(c)

Let $a, b, c$ be the sides of a triangle.

$	herefore \quad a^2+b^2 \geq 2 a b \quad[\because AM \geq GM ]$

Similarly, $b^2+c^2 \geq 2 b

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) Let $a, b, c$ be the sides of a triangle. $ herefore \quad a^2+b^2 \geq 2 a b \quad[\because AM \geq GM ]$ Similarly, $b^2+c^2 \geq 2 b c$ $c^2+a^2 \geq 2 a c$ $\Rightarrow 2\left(a^2+b^2+c^2 ight) \geq 2(a b+b c+c a)$ $\Rightarrow \quad \frac{a^2+b^2+c^2}{a b+b c+a c} \geq 1$ $t \geq 1 \quad\left[\because \frac{a^2+b^2+c^2}{a b+b c+a c}=t ight]$ $\because a, b, c$ 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 $a^2+b^2+c^2 < 2(a b+b c+c a)$ $\Rightarrow \quad \frac{a^2+b^2+c^2}{a b+b c+c a} < 2$ $ herefore \quad t < 2$ $ herefore \quad\{x \in R \mid 1 \leq x < 2\}$

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

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