α, β, γ are real and satisfy α2 + β2 + γ2 = 1. Show that -1/2 ≤ αβ + βγ + γα ≤ 1.
Solution
We have (α + β + γ)2 ≥ 0, so α2 + β2 + γ2 ≥ -2(αβ + βγ + γα), so -½ ≤ αβ + βγ + γα.
Also (α - β)2 + (β - γ)2 + (γ + α)2 ≥ 0, so 2(α2 + β2 + γ2) ≥ 2(αβ + βγ + γα), so αβ + βγ + γα ≤ 1.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03