Let a, b and c be three positive real numbers | vedclass

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Question

(C) For the roots of the quadratic equation $x^2+2(a+b+c)x+3\lambda(ab+bc+ca)=0$ to be real,the discriminant $D$ must be greater than or equal to $0$.

Vedclass · Accepted Answer

$\lambda < \frac{4}{3}$

Vedclass · Answer

(C) For the roots of the quadratic equation $x^2+2(a+b+c)x+3\lambda(ab+bc+ca)=0$ to be real,the discriminant $D$ must be greater than or equal to $0$.$D = [2(a+b+c)]^2 - 4(1)[3\lambda(ab+bc+ca)] \geq 0$$4(a+b+c)^2 - 12\lambda(ab+bc+ca) \geq 0$$(a+b+c)^2 - 3\lambda(ab+bc+ca) \geq 0$$a^2+b^2+c^2+2(ab+bc+ca) - 3\lambda(ab+bc+ca) \geq 0$$a^2+b^2+c^2 \geq (3\lambda-2)(ab+bc+ca)$$\lambda \leq \frac{a^2+b^2+c^2}{3(ab+bc+ca)} + \frac{2}{3}$Since $a, b, c$ are sides of a triangle,we have $a+b > c$,$b+c > a$,and $c+a > b$. This implies $(a+b-c)(b+c-a)(c+a-b) > 0$,which leads to the inequality $a^2+b^2+c^2 Dividing by $(ab+bc+ca)$,we get $\frac{a^2+b^2+c^2}{ab+bc+ca} Substituting this into the expression for $\lambda$:$\lambda < \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$.

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

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