www.whkt.net > 如何证明:X2+Y2+Z2(均为平方) 〉=2XYCosA+2YZCos...

如何证明:X2+Y2+Z2(均为平方) 〉=2XYCosA+2YZCos...

直接用余弦定理就可以了再看看别人怎么说的.

x2+y2+z2≥2xycosC+2yzcosA+2zxcosB<=>x+y+z-x(2ycosC+2zcosB)-2yzcosA≥0<=>(x-ycosC-zcosB)+y+z-2yzcosA-(ycosC+zcosB)≥0<=>(x-ycosC-zcosB)+ysinC+zsinB-yz(2cosA+2cosCcosB)≥0<=>(x-ycosC-zcosB)+y

方法一、x2+y2+z2≥2xycosC+2yzcosA+2zxcosBx+y+z-x(2ycosC+2zcosB)-2yzcosA≥0(x-ycosC-zcosB)+y+z-2yzcosA-(ycosC+zcosB)≥0(x-ycosC-zcosB)+ysinC+

x2+y2+z2=xy+yz+zx即2x2+2y2+2z2=2xy+2yz+2zx即2x2+2y2+2z2-2xy-2yz-2zx=0即(X-Y)^2+(Y-Z)^2+(Z-X)^2=0所以x=y=z

2xycosC+2yzcosB+2xzcosA=x^2+y^2-z^2+(y^2+z^2-x^2)+(x^2+z^2-y^2)=x^2+y^2+z^2

证明:∵x^2+y^2≥2xyy^2+z^2≥2yzx^2+z^2≥2xz三项相加:∴2(x^2+y^2+z^2)≥2*(xy+yz+xz)∴x^2+y^2+z^2≥xy+yz+xz

这个貌似还真难住我了~有一定思路~写出来大家一起探讨一下吧~x方+y方》=2xycosA;x方+z方》=2xzcosB;y方+z方》=2yzcosC;因此,相加,就只需要证明cosA+cosB+cosC》=2(cosA+cosB+cosC)也就是说cosA+cosB+cosC>=0.5后面究竟应该怎样利用A+B+C=180度证明我还真不太了解.

利用余弦公式:z^2=x^2+y^2-2xycosC.可以化简为2z^2>=2z(ycosA+xcosB)消掉公因子,利用三角性质显然成立.得证.

因为(x-y)≥0所以x+y≥2xy同理y+z≥2yzx+z≥2xz将以上三式相加得到:2(x+y+z)≥2(xy+yz+xz)即x+y+z≥xy+yz+xz

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