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
$x^{2}+2(a+b+c) x+3 \lambda(a b+b c+c a)=0$ are real, are given by

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

Answer 1

Answer: (C)

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