a, b, c are positive reals. Let a' = √(a^{2} + 8bc), b' = √(b^{2} + 8ca), c' = √(c^{2} + 8ab). Prove that a/a' + b/b' + c/c' >= 1.

**Solution**

A not particularly elegant, but fairly easy, solution is to use Cauchy: (∑ xy)^{2} ≤ ∑ x^{2} ∑ y^{2}.

To get the inequality the right way around we need to take x^{2} = a/a' [to be precise, we are taking x_{1}^{2} = a/a', x_{2}^{2} = b/b', x_{3}^{2} = c/c'.]. Take y^{2} = a a', so that xy = a. Then we get ∑ a/a' >= (∑ a)^{2}/∑ a a'.

Evidently we need to apply Cauchy again to deal with ∑ a a'. This time we want ∑ a a' ≤ something. The obvious X=a, Y=a' does not work, but if we put X=a^{1/2}, Y=a^{1/2}a', then we have ∑ a a' ≤ (∑ a)^{1/2} (∑ a a'^{2})^{1/2}. So we get the required inequality provided that (∑ a)^{3/2} ≥ (∑ a a'^{2})^{1/2} or (∑ a)^{3} ≥ ∑ a a'^{2}.

Multiplying out, this is equivalent to: 3(ab^{2} + ac^{2} + ba^{2} + bc^{2} + ca^{2} + cb^{2}) ≥ 18abc, or a(b - c)^{2} + b(c - a)^{2} + c(a - b)^{2} ≥ 0, which is clearly true.

© John Scholes

jscholes@kalva.demon.co.uk

12 Aug 2001

Last corrected/updated 12 Aug 2001