如题所述
a2+b2+c2-ab-ac-bc
=1/2(2a2+2b2+2c2-2ab-2ac-2bc)
=1/2[(a-b)^2+(b-c)^2+(a-c)^2]
å 为ï¼(a-b)^2>=0,(b-c)^2>=0,(a-c)^2>=0
æ以ï¼
a2+b2+c2-ab-ac-bc=1/2[(a-b)^2+(b-c)^2+(a-c)^2]>=0
a2+b2+c2-ab-ac-bcæ¯éè´æ°
=1/2(2a2+2b2+2c2-2ab-2ac-2bc)
=1/2[(a-b)^2+(b-c)^2+(a-c)^2]
å 为ï¼(a-b)^2>=0,(b-c)^2>=0,(a-c)^2>=0
æ以ï¼
a2+b2+c2-ab-ac-bc=1/2[(a-b)^2+(b-c)^2+(a-c)^2]>=0
a2+b2+c2-ab-ac-bcæ¯éè´æ°
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