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)