已知a^2+b^2+c^2=1,求证-1/2<=ab+bc+ca<=1

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已知a^2+b^2+c^2=1,求证-1/2<=ab+bc+ca<=1已知a^2+b^2+c^2=1,求证-1/2<=ab+bc+ca<=1已知a^2+b^2+c^2=1,求证-1/2<=ab+bc+

已知a^2+b^2+c^2=1,求证-1/2<=ab+bc+ca<=1
已知a^2+b^2+c^2=1,求证-1/2<=ab+bc+ca<=1

已知a^2+b^2+c^2=1,求证-1/2<=ab+bc+ca<=1
2(a2+b2+c2)-2(ab+bc+ca)=(a-b)2+(b-c)2+(c-a)2≥0
所以:2(a2+b2+c2)≥2(ab+bc+ca)
即:ab+bc+ca≤a2+b2+c2=1
同理:2(a2+b2+c2)+2(ab+bc+ca)=(a+b)2+(b+c)2+(c+a)2≥0
即:ab+bc+ca≥-(a2+b2+c2)=-1
所以:最大值为1,最小值为-1

2(a^2+b^2+c^2)+2ab+2bc+2ac=2+2ab+2bc+2ac=(a+b)^2+(b+c)^2+(c+a)^2>=0
得-1/2<=ab+bc+ca
2(a^2+b^2+c^2)-(2ab+2bc+2ac)=2-(2ab+2bc+2ac)=(a-b)^2+(b-c)^2+(c-a)^2>=0
得ab+bc+ca<=1