If are the sides of a triangle such that has real roots, then
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
For real roots, discriminant ≥ 0:
4(a+b+c)² - 12λ(ab+bc+ca) ≥ 0
→ λ ≤ (a+b+c)² / 3(ab+bc+ca)
Now, (a+b+c)² = a²+b²+c² + 2(ab+bc+ca) ≥ 3(ab+bc+ca) (since a²+b²+c² ≥ ab+bc+ca), so the ratio is ≥ 1. But we need the UPPER bound on this ratio.
For a triangle, we use the known inequality: (a+b+c)² < 4(ab+bc+ca). This comes from the triangle inequalities: since a < b+c, b < a+c, c < a+b, squaring and adding gives a²+b²+c² < 2(ab+bc+ca), so (a+b+c)² = a²+b²+c²+2(ab+bc+ca) < 4(ab+bc+ca).
Therefore (a+b+c)²/3(ab+bc+ca) < 4/3.
So λ ≤ (a+b+c)²/3(ab+bc+ca) < 4/3, which means λ < 4/3 is the necessary condition for the equation to have real roots for some valid triangle.
4(a+b+c)² - 12λ(ab+bc+ca) ≥ 0
→ λ ≤ (a+b+c)² / 3(ab+bc+ca)
Now, (a+b+c)² = a²+b²+c² + 2(ab+bc+ca) ≥ 3(ab+bc+ca) (since a²+b²+c² ≥ ab+bc+ca), so the ratio is ≥ 1. But we need the UPPER bound on this ratio.
For a triangle, we use the known inequality: (a+b+c)² < 4(ab+bc+ca). This comes from the triangle inequalities: since a < b+c, b < a+c, c < a+b, squaring and adding gives a²+b²+c² < 2(ab+bc+ca), so (a+b+c)² = a²+b²+c²+2(ab+bc+ca) < 4(ab+bc+ca).
Therefore (a+b+c)²/3(ab+bc+ca) < 4/3.
So λ ≤ (a+b+c)²/3(ab+bc+ca) < 4/3, which means λ < 4/3 is the necessary condition for the equation to have real roots for some valid triangle.
[AIEEE 2006]