求证：x^2+y^2+z^2>=2xycos(A)+2xzcos(B)+2yzcos(C),ABC为三角形x,y,z属于实数

Question

求证：x^2+y^2+z^2>=2xycos(A)+2xzcos(B)+2yzcos(C),ABC为三角形
x,y,z属于实数

Answer 1

方法一、x2+y2+z2≥2xycosC+2yzcosA+2zxcosB
x²+y²+z²-x(2ycosC+2zcosB)-2yzcosA≥0
(x-ycosC-zcosB)²+y²+z²-2yzcosA-(ycosC+zcosB)²≥0
(x-ycosC-zcosB)²+y²sin²C+z²sin²B-yz(2cosA+2cosCcosB)≥0
(x-ycosC-zcosB)²+y²sin²C+z²sin²B-yz[-2cos(B+C)+2cosCcosB]≥0
(x-ycosC-zcosB)²+y²sin²C+z²sin²B-yz[-2cosCcosB+2sinBsinC+2cosCcosB]≥0
(x-ycosC-zcosB)²+y²sin²C+z²sin²B-yz2sinBsinC≥0
(x-ycosC-zcosB)²+(ysinC-zsinB)²≥0
上不等式显然成立,故原命题成立
当x=ycosC+zcosB,ysinC=zsinB时取等号
方法二、
因为,cosA=(x^2+y^2-z^2)/2xy,所以,2xycosA=x^2+y^2-z^2
同理,可得2xzcosB=x^2+z^2-y^2,
2yzcosc=y^2+z^2-x^2
然后代进去就好了,就可以得出答案了.