Let a, b and c be the sides of a scalene tria | vedclass

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(A) Given the quadratic equation $x^2+2(a+b+c)x+3\lambda(ab+bc+ca)=0$ has real roots,the discriminant $D \geq 0$.$D = [2(a+b+c)]^2 - 4(1)(3\lambda(ab+

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$\left(-\infty, \frac{4}{3}\right)$

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(A) Given the quadratic equation $x^2+2(a+b+c)x+3\lambda(ab+bc+ca)=0$ has real roots,the discriminant $D \geq 0$.$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 \geq 3\lambda(ab+bc+ca)$$\lambda \leq \frac{(a+b+c)^2}{3(ab+bc+ca)}$For a scalene triangle,we know that $(a-b)^2 + (b-c)^2 + (c-a)^2 > 0$.Expanding this,$2(a^2+b^2+c^2) - 2(ab+bc+ca) > 0$,so $a^2+b^2+c^2 > ab+bc+ca$.Adding $2(ab+bc+ca)$ to both sides,we get $(a+b+c)^2 > 3(ab+bc+ca)$.Thus,$\frac{(a+b+c)^2}{3(ab+bc+ca)} > 1$.Also,from the triangle inequality,$(a+b+c)^2 Since $\lambda \leq \frac{(a+b+c)^2}{3(ab+bc+ca)}$ and the expression is strictly less than $\frac{4}{3}$,the range for $\lambda$ is $\left(-\infty, \frac{4}{3}\right)$.

Let $a, b$ and $c$ be the sides of a scalene triangle. If $\lambda$ is a real number such that the roots of the equation $x^2+2(a+b+c)x+3\lambda(ab+bc+ca)=0$ are real,then the interval in which $\lambda$ lies is

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