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If $a,b,c$ are sides of a triangle $ABC$ such that $x^2-2(a+b+c)x+3 \lambda (ab+bc+ca)=0$ has real roots then

  1. $\lambda < \frac{4}{3}$
  2. $\lambda > \frac{5}{3}$
  3. $\lambda \in \big( \frac{4}{3}, \frac{5}{3}\big)$
  4. $\lambda \in \big( \frac{1}{3}, \frac{5}{3}\big)$
in Numerical Ability by Veteran (425k points)
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1 Answer

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Answer: $A$

The discriminant, $D$
$$ = 4(a+b+c)^2-12\lambda(ab+bc+ca) \geq 0$$
$$=\frac{(a+b+c)^2}{ab+bc+ca} \geq3\lambda \qquad \to (1)$$

Now, $$ \because (a+b+c)^2 = a^2 +b^2 + c^2 + 2ab +2bc +2ca=a^2 + b^2 + c^2 +2(ab+bc+ca)$$Substituting this value in $(1)$, we get:
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+2\geq3\lambda$$
Now, we know that the sum of any two sides of a $\triangle $ is greater than the third side:
$$\therefore a < b + c \implies(a-c) < b \implies (a-c)^2 < b^2 \implies a^2 + b^2 -2ac < b^2 \qquad \to (2)$$
Similarly,
$$a < b + c \implies (a-b)^2 < c^2 \implies a^2 + b^2 -2ab < c^2 \qquad \to(3)$$
$$b < a + c \implies (b-c)^2 < a^2 \implies  b^2 + c^2 -2bc < a^2 \qquad \to (4)$$
Adding equations $(2)$, $(3)$ and $(4)$, we get:
$$2 > \frac{a^2+b^2+c^2}{ab+bc+ca}$$
$$\therefore 2 + 2 > 3\lambda \implies \lambda < \frac{4}{3}$$
$\therefore \; A$ is the correct option.
by Boss (13.5k points)
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