Mathematics - Properties of Triangles Question with Solution | TestHub

Since, roots are real, $D\ge 0$ $$\begin{array}{ll}\Rightarrow & 4(a+b+c{)}^2-12\lambda (ab+bc+ca)\ge 0\\ \Rightarrow & (a+b+c{)}^2\ge 3\lambda (ab+bc+ca)\\ \Rightarrow & a^2+b^2+c^2\ge (ab+bc+ca)(3\lambda -2)\\ \Rightarrow & 3\lambda -2\le \frac{a^2+b^2+c^2}{ab+bc+ca}\\ \text{ where, }{a}^2+b^2\ge 2ab,b^2+c^2\ge 2bc\text{ and }{c}^2+a^2\ge 2ca& \\ \Rightarrow & a^2+b^2+c^2\ge (ab+bc+ca)\\ \therefore & \frac{a^2+b^2+c^2}{ab+bc+ca}\ge 1\end{array}$$ Also, $\cos A=\frac{b^2+c^2-a^2}{2bc}<1$ $$\Rightarrow b^2+c^2-a^2<2bc$$ Similarly, $c^2+a^2-b^2<2ca$ and $a^2+b^2-c^2<2ab$ $$\Rightarrow a^2+b^2+c^2<2(ab+bc+ca)\text{ or }\frac{a^2+b^2+c^2}{ab+bc+ca}<2$$ From Eqs. (i), (ii) and (iii) $$\begin{array}{lrl}\therefore & 3\lambda -2& <2\\ \Rightarrow & \lambda <\frac{4}{3}& \end{array}$$