$\begin{array}{l} L e t a, b, c b e t h e t h r e e p o s i t i v e r e a l n u m b e r s s u c h t h a t t h e s u m o f a n y t w o o f t h e m i s g r e a t e r t h a n t h e t h i r d . A l l t h e v a l u e s o f λ s u c h t h a t t h e r o o t s o f \\ t h e e q u a t i o n x^{2} + 2 (a + b + c) x + 3 λ (a b + b c + c a) = 0 a r e r e a l a r e g i v e n b y \end{array}$

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$λ \geq \frac{2}{3}$

$λ < \frac{2}{3}$

$λ < \frac{4}{3}$

$\frac{1}{3} < λ < \frac{2}{3}$

answer is C.

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Detailed Solution

$\begin{array}{l} G i v e n s u m o f t w o r o o t s i s g r e a t e r t h a n t h i r d \\ ∴ a < b + c \Rightarrow a - c < b \\ \Rightarrow {(a - c)}^{2} < b^{2} \\ \Rightarrow a^{2} + c^{2} - 2 a c < b^{2} - - (i) \\ s^{l y} a^{2} + b^{2} - 2 a b < c^{2} - - (i i) \\ b^{2} + c^{2} - 2 b c < a^{2} - - (i i i) \\ A d d i n g (i), (i i), (i i i) \Rightarrow a^{2} + b^{2} + c^{2} < 2 (a b + b c + c a) \\ \Rightarrow \frac{a^{2} + b^{2} + c^{2}}{a b + b c + c a} < 2 - - (A) \\ r o o t s o f x^{2} + 2 (a + b + c) x + 3 λ (a b + b c + c a) = 0 \\ a r e r e a l \Rightarrow D \geq 0 \\ \Rightarrow 4 {(a + b + c)}^{2} - 4 (3 λ) (a b + b c + c a) \geq 0 \\ \Rightarrow a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) \geq 3 λ (a b + b c + c a) \\ \Rightarrow λ \leq \frac{a^{2} + b^{2} + c^{2}}{3 (a b + b c + c a)} + \frac{2}{3} - - (B) \\ f r o m A a n d B \\ λ < \frac{2}{3} + \frac{2}{3} \\ \Rightarrow λ < \frac{4}{3} \end{array}$