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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($1$) Let $E_1, E_2$ and $F_1 F_2$ be the chords of $S$ passing through the point $P_0(1,1)$ and parallel to the $x$-axis and the $y$-axis, respectively. Let $G _1 G _2$ be the chord of $S$ passing through $P _0$ and having slope -$1$ . Let the tangents to $S$ at $E_1$ and $E_2$ meet at $E_3$, the tangents to $S$ at $F_1$ and $F_2$ meet at $F_3$, and the tangents to $S$ at $G_1$ and $G_2$ meet at $G_3$. Then, the points $E_3, F_3$, and $G _3$ lie on the curve
$(A)$ $x+y=4$ $(B)$ $(x-4)^2+(y-4)^2=16$ $(C)$ $(x-4)(y-4)=4$ $(D)$ $x y=4$
($2$) Let $P$ be a point on the circle $S$ with both coordinates being positive. Let the tangent to $S$ at $P$ intersect the coordinate axes at the points $M$ and $N$. Then, the mid-point of the line segment MN must lie on the curve
$(A)$ $(x+y)^2=3 x y$ $(B)$ $x^{2 / 3}+y^{2 / 3}=2^{4 / 3}$ $(C)$ $x^2+y^2=2 x y$ $(D)$ $x^2+y^2=x^2 y^2$
Give the answer or quetion ($1$) and ($2$)