Question

Let a, b, c be the sides of a triangle. no two of them are equal and λ ∈ r. if the roots of the equation x2 + 2(a + b+

c.x + 3λ (ab + bc + ca) = 0 are real, then

Answer 1

I hope u can try further

Answer 2

we know that difference of two sides of a triangle is less than the third side.

$|a - b| < c \: \: \: \: \: \: \: \: {a}^{2} + {b}^{2} - 2ab < {c}^{2} \: ....(1) \\ |b - c| < a \: \: \: \: \: \: \: \: {b}^{2} + {c}^{2} - 2bc < {a}^{2} \: ....(2) \\ |c - a| < b \: \: \: \: \: \: \: \: {c}^{2} + {a}^{2} - 2ca < {b}^{2} \: ....(3)$

Now adding equation (1), (2), (3)

${a}^{2} + {b}^{2} + {c}^{2} < 2(ab + bc + ca) \\ \frac{ {a}^{2} + {b}^{2} + {c}^{2} }{ab + bc + ca} < 2 \: ....(4)$

we know that if roots are real then (∆) ≥ 0

→ b² - 4ac ≥ 0

4(a²+b²+c²+2ab+2bc+2ca) - 12λ(ab+bc+ca) ≥ 0

a²+b²+c²+2ab+2bc+2ca ≥ 3λ(ab+bc+ca)

a² + b² + c² ≥ (3λ - 2)(ab + bc + ca)

→ a² + b² + c² / (ab + bc + ca) ≥ 3λ - 2 .... (5)

→ From equation (4), (5)

2ab + 2bc + 2ca > (3λ - 2)(ab + bc + ca)

→ 2 > 3λ - 2 → λ < 4/3

Hope it's helpful