Question
Let $$a, b, c$$ be the sides of a triangle where $$a \ne b \ne c\,\,{\text{and}}\,\,\lambda \in R.$$ If the roots of the equation $${x^2} + 2\left( {a + b + c} \right)x + 3\lambda \left( {ab + bc + ca} \right) = 0$$ are real, then
A.
$$\lambda < \frac{4}{3}$$
B.
$$\lambda > \frac{5}{3}$$
C.
$$\lambda \in \left( {\frac{1}{3},\frac{5}{3}} \right)$$
D.
$$\lambda \in \left( {\frac{4}{3},\frac{5}{3}} \right)$$
Answer :
$$\lambda < \frac{4}{3}$$
Solution :
$$\because \,\,a,b,c$$ are sides of a triangle and $$a \ne b \ne c$$
$$\therefore \,\,\left| {a - b} \right| < \left| c \right|$$
$$\eqalign{
& \Rightarrow \,\,{a^2} + {b^2} - 2ab < {c^2} \cr
& {\text{Similarly, we get}} \cr
& {b^2} + {c^2} - 2bc < {a^2};{c^2} + {a^2} - 2ca < {b^2} \cr
& {\text{On adding, we get}} \cr
& {a^2} + {b^2} + {c^2} < 2\left( {ab + bc + ca} \right) \cr
& \Rightarrow \,\,\frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}} < 2\,\,\,\,\,\,\,.....\left( 1 \right) \cr} $$
$$\because $$ Roots of the given equation are real
$$\eqalign{
& \therefore \,\,{\left( {a + b + c} \right)^2} - 3\lambda \left( {ab + bc + ca} \right) \geqslant 0 \cr
& \Rightarrow \,\,\frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}} \geqslant 3\lambda - 2\,\,\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr
& {\text{From}}\left( 1 \right){\text{and}}\left( 2 \right),{\text{we get}} \cr
& {\text{3}}\lambda - 2 < 2 \cr
& \Rightarrow \,\,\lambda < \frac{4}{3}. \cr} $$