a, b, c are positive reals. Let a' = √(a2 + 8bc), b' = √(b2 + 8ca), c' = √(c2 + 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 ≤ ∑ x2 ∑ y2.
To get the inequality the right way around we need to take x2 = a/a' [to be precise, we are taking x12 = a/a', x22 = b/b', x32 = c/c'.]. Take y2 = 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=a1/2, Y=a1/2a', 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(ab2 + ac2 + ba2 + bc2 + ca2 + cb2) ≥ 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