Question

Let a, b, c be the sides of a triangle, no two of them are equal and $\lambda \in R$. If the roots of the equation $x^2+2(a+b+c)x+3\lambda$( ab + bc +ca ) = 0 are real, then

• A. <
• B. >
• C.
• D.

Answer 1

The correct option is A

<

$x^2+2(a+b+c)x+3\lambda$ (ab+bc+ ca) = 0

since roots are real, D $\ge$ 0

$\Rightarrow$ $4(a+b+c{)}^2-4.3\lambda$ (ab+bc+ca) $\ge$ 0

$\Rightarrow$ $(a+b+c{)}^2-3\lambda (ab+bc+ca)\ge$ 0

$\Rightarrow$ $a^2+b^2+c^2\ge (3\lambda -2)$ (ab+bc+ ca)
$\Rightarrow \frac{a^2+b^2+c^2}{(ab+bc+ca)}\ge (3\lambda -2)\cdots \cdots \left(i\right)$
We know that in $△$ ABC , cos A = $\frac{b^2+c^2-a^2}{2bc}$ <1

$\Rightarrow$ $b^2+c^2-a^2$ < $2bc$ ....(ii)

Similarly $\Rightarrow$ $c^2+a^2-b^2$< $2ca$....(iii) and $\Rightarrow$ $a^2+b^2-c^2$<$2ab$ ...(iv)

Adding all these we get $\Rightarrow$ $a^2+b^2+c^2$ < 2 (ab + bc + ca )

$\Rightarrow$ $\frac{a^2+b^2+c^2}{ab+bc+ca}$ < 2

Hence from (i) we have $(3\lambda -2)<2$ $\Rightarrow$ $\lambda$ < $\frac{4}{3}$