已知：a^2+b^2+c^2=1.a,b,c为实数，求证：－１／２＜＝ab+ac+bc<=1.
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1 = a^2+b^2+c^2 =[(a^2+b^2)+(b^2+c^2)+(c^2+a^2)]/2= [2ab+2bc+2ca]/2 = ab+bc+ca=== ab+bc+ca = -1=== ab+bc+ca = -1/2因此：-1/2 <= ab+bc+ca <= 1
已知：a^2+b^2+c^2=1.a,b,c为实数，求证：－１／２＜＝ab+ac+bc<=1.
1 = a^2+b^2+c^2 =[(a^2+b^2)+(b^2+c^2)+(c^2+a^2)]/2= [2ab+2bc+2ca]/2 = ab+bc+ca=== ab+bc+ca = -1=== ab+bc+ca = -1/2因此：-1/2 <= ab+bc+ca <= 1