IIT JEE Advanced 2006 Paper 1

We have
x2 – 2(a + b + c)x + 3λ(ab + bc + ca) = 0
D ≥ 0
4(a+b+c)2 − 12λ(ab + bc + ca) ≥ 0
(a+b+c)2 − 3λ(ab + bc + ca) ≥ 0
a2+b2+c2 + (ab + bc + ca)(2 − 3λ) ≥ 0
Therefore,
a2+b2+c2 ≥ (3λ - 2)(ab + bc + ca)
3λ - 2 ≤

a2+b2+c2

ab+bc+ca

λ ≤

1

3

(

a2+b2+c2

ab+bc+ca

)+

2

3

and we also have |a – b|< c, |b – c| < a and |c – a| < b. Therefore,
a2+b2 – 2ab < c2 (1)
a2+b2–c2 < 2ab
Similarly, we get
b2+c2–a2 < 2bc (2)
c2+a2–b2 < 2ca (3)
Adding Eqs. (1), (2) and (3), we get
a2+b2+c2 < 2(ab + bc + ca)

a2+b2+c2

ab+bc+ca

< 2
Then λ <

2

3

+

2

3

⇒ λ=

4

3