Let a, b, c be the sides of a triangle where $a\neq c$ and $k \in R$. If the roots of the equation $x^2+ 2(a + b +c)x + 3k(ab + bc + ca) = 0$ are real, then find the interval in which $k$ lies.
I have used the fact that equation has real roots, but how to use the fact a,b,c, are sides of a triangle.
On
Since $a,b,c$ are sides of a triangle,
$$|a-b|<c, \quad |b-c|<a, \quad |c-a|<b$$
Squaring and adding these inequalities, we get
$$a^2+b^2+c^2-2ab-2bc-2ca < 0$$
i.e. $$\dfrac{a^2+b^2+c^2}{ab+bc+ca} < 2$$
From the discriminant equation for the quadratic, we have $$k \leq \dfrac{a^2+b^2+c^2}{3(ab+bc+ca)} + \dfrac{2}{3}$$
and hence $$k < \dfrac{4}{3}$$
$$(2(a+b+c))^2\ge4(1)(3k(ab+bc+ca)$$ (since if $ax^2+bx+c=0$ then $b^2\ge4(a)(c)$ for real roots to exist) $$k\le\frac{4(a+b+c)^2}{12(ab+bc+ca)}$$