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

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Aicharya · Accepted Answer

Roots of the equation $ x^2+2(a+b+c) x+3 \lambda(a b+b c+c a)=0 $ are real, so $ \begin{aligned} & {[2(a+b+c)]^2-4(1)[3 \lambda(a b+b c+c a)] \geq 0} \\ & \Rightarrow \quad \lambda \leq \frac{a^2+b^2+c^2+2(a b+b c+c a)}{3(a b+b c+c a)} \end{aligned} $

Now, sum of two roots is greater than thirds So, $ \begin{aligned} & a < b+c \Rightarrow(a-c) < b \\ & \Rightarrow \quad(a-c)^2 < b^2 \\ & \end{aligned} $ $\Rightarrow \quad a^2+c^2-2 a c < b^2$

Adding Eqs. (ii), (iii) and (iv), we get $ \begin{aligned} & a^2+b^2+c^2 < 2(a b+b c+c a) \\ \Rightarrow \quad & \frac{a^2+b^2+c^2}{a b+b c+c a} < 2 \end{aligned} $

$\lambda < \frac{2}{3}+\frac{2}{3} \Rightarrow \lambda < \frac{4}{3}$

Let $a, b$ and $c$ be three positive real numbers such that the sum of any two of them is greater than the…

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