Question

Let a, b, c be the sides of the 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

$D\ge 0\Rightarrow 4(a+b+c{)}^2-12\lambda (ab+bc+ca)\ge 0$
$\Rightarrow \lambda \le \frac{a^2+b^2+c^2}{3(ab+bc+ca)}+\frac{2}{3}......\left(1\right)$
Since $|a-b|<c\Rightarrow a^2+b^2-2ab<c^2......\left(2\right)$
$|b-c|<a\Rightarrow b^2+c^2-2bc<a^2......\left(3\right)$
$|c-a|<b\Rightarrow c^2+a^2-2ca<b^2.......\left(4\right)$
Adding (2),(3) and (4)
$a^2+b^2+c^2-2(ab+bc+ca)<0$
$\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca}<2.$
$From\left(1\right),\lambda <\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$.