Question

Let $a,b$ and $c$ be three positive real numbers such that the sum of any two of them is greater than the third. All the values of $\lambda$ such that the roots of the equation
$x^2+2(a+b+c)x+3\lambda (ab+bc+ca)=0$ are real, are given by

• A. $\lambda <\frac{2}{3}$
• B. $\lambda \ge \frac{2}{3}$
• C. $\lambda <\frac{4}{3}$
• D. $\frac{1}{3}<\lambda <\frac{2}{3}$

Answer 1

Answer: (C)

Roots of the equation
$x^2+2(a+b+c)x+3\lambda (ab+bc+ca)=0$
are real, so $[2(a+b+c){]}^2-4(1)[3\lambda (ab+bc+ca)]\ge 0$
$\Rightarrow \lambda \le \frac{a^2+b^2+c^2+2(ab+bc+ca)}{3(ab+bc+ca)}$
$\Rightarrow \lambda \le \frac{a^2+b^2+c^2}{3(ab+bc+ca)}+\frac{2}{3}...(i)$
Now, sum of two roots is greater than thirds
So, $a<b+c\Rightarrow (a-c)<b$
$\Rightarrow (a-c{)}^2<b^2$
$\Rightarrow a^2+c^2-2ac<b^2...(ii)$
Similarly, $a^2+b^2-2ab<c^2...(iii)$
and $c^2+b^2-2bc<a^2...(iv)$
Adding Eqs. (ii), (iii) and (iv), we get
$a^2+b^2+c^2<2(ab+bc+ca)$
$\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca}<2$
$\Rightarrow 2>\frac{a^2+b^2+c^2}{ab+bc+ca}...(v)$
[from Eqs. (i) and (v)]
$\lambda <\frac{2}{3}+\frac{2}{3}$
$\Rightarrow \lambda <\frac{4}{3}$