MHB Value of lambda in quadratic equation

[juantheron]

If $a,b,c$ are the length of the sides of an scalene triangle, If the equation

$x^2+2(a+b+c)x+3\lambda\left(ab+bc+ca\right) = 0$ has real and distinct roots,

Then the value of $\lambda$ is given by

Options::

(a) $\displaystyle \lambda < \frac{4}{3}\;\;\;\;\; $ (b)$\displaystyle \frac{11}{3}< \lambda < \frac{17}{3}\;\;\;\;\;\;$ (c) $ \lambda \geq 1\;\;\;\;\;\;$ (d)$ \displaystyle \frac{11}{3}< \lambda < \frac{17}{3}$

My Try:: Given that $\triangle$ is scalene means $a+b>c$ and $b+c>a$ and $c+a>b$

and given equation has real and distinct roots, Then $D>0$

So $\displaystyle 4(a+b+c)^2-4\cdot 3 \lambda \left(ab+bc+ca\right)>0$

So $(a+b+c)^2 - 3\lambda \left(ab+bc+ca\right)>0\Rightarrow (a^2+b^2+c^2)-(3\lambda-2)\cdot (ab+bc+ca)>0$

So $\displaystyle (a^2+b^2+c^2)>(3\lambda-2)(ab+bc+ca)>0$

So $\displaystyle 3\lambda-2 < \frac{a^2+b^2+c^2}{ab+bc+ca}\Rightarrow 3\lambda < \frac{(a+b+c)^2}{ab+bc+ca}$

Now How can I solve after that, Help me

Thanks

 

[Opalg]

You have done most of the work by getting to the inequality $$3\lambda-2 < \frac{a^2+b^2+c^2}{ab+bc+ca}$$. What remains is to make use of the scalene conditions $a+b>c$, $b+c>a$ and $c+a>b$. I suggest you multiply the first one by $c$, the second one by $a$, and the third one by $b$. Then add them. That should give you an estimate for the fraction $ \frac{a^2+b^2+c^2}{ab+bc+ca}.$

 

[juantheron]

Thanks opalg Got it.

Using yours hint::

$\displaystyle (a+b)>c\Rightarrow \left(ac+bc\right)>c^2$

$\displaystyle (b+c)>a\Rightarrow \left(ab+ac\right)>a^2$

$\displaystyle (c+a)>b\Rightarrow \left(bc+ab\right)>b^2$

Now add All These equation, we got

$\displaystyle \Rightarrow 2(ab+bc+ca) > (a^2+b^2+c^2)\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca} < 2$

So $\displaystyle 3\lambda - 2 < 2\Rightarrow \lambda < \frac{4}{3}$