# a, b, c > 0. Prove that 1/(a² + ab + b²) + 1/(b² + bc + c²) + 1/(c² + ca + a²) ≥ 9/(a + b + c)².?

### 2 Answers

- IndicaLv 75 years agoFavourite answer
This hard inequality seems to be quite well-known (especially in Viet Nam) and I did discover quite a

few published solutions, notably one by Vo Quoc Ba Can in his book “Inequalities with Beautiful

Solutions” which I’ve referred to before. Apparently, it can also be done by clearing denominators

and applying basics. However the solution that I liked is this because, given the time, I think it’s the

most likely to be discovered by non-specialists.

(a+b+c)² ∑ 1/(b²+c²+bc) ≥ 9 is homogeneous so you can write a+b+c=1 → ∑1/(b²+c²+bc) ≥ 9

b²+c²+bc = (∑a)²−a²−bc−2ac−2ab = 1−∑ab−a∑a = 1−∑ab−a = 1−x−a where x=∑ab

Inequality is ∑1/(1−x−a) ≥ 9 → ∑(1−x−a)(1−x−b) ≥ 9(1−x−a)(1−x−b)(1−x−c)

→ 3(1−x)²−2(∑a)(1−x)+∑ab ≥ 9( (1−x)³−(∑a)(1−x)²+(∑ab)(1−x)−abc )

→ 3(1−x)²−2(1−x)+x ≥ 9( (1−x)³−(1−x)²+x(1−x)−abc )

This simplifies to 3x(1+2x−3x²) ≤ 1+9abc

Now 1+2x−3x² = 4/3–3(x−⅓)² ≤ 4/3 so we require 4∑ab ≤ 1+9abc … (i)

(i) is just Schur with t=1

∑ a(a−b)(a−c) = ∑a(a²−ab−ac+bc) = ∑a( a²−∑ab+2bc) = ∑a³ − ∑a∑ab + 6abc ≥ 0

= { (∑a)³−3∑a∑ab+3abc } − ∑a∑ab + 6abc = (∑a)³ − 4∑a∑ab + 9abc ≥ 0

Setting ∑a=1 gives 4∑ab ≤ 1 + 9abc