不等式问题4在非钝角△ABC中,
我们知道sin^2A=1/(1+cot^2A)
令x=cotA,y=cotB,z=cotC,可以得到xy+yz+zx=1,
原不等式化为
1/(x+y)^2+1/(y+z)^2+1/(z+x)^2≥9/4 (*)
先看一个变形
1/(x+y)^2=(xy+yz+zx)/(x+y)^2=xy/(x+y)^2+z/(x+y)
所以(*)右边
=x/(y+z)+y/(z+x)+z/(x+y)+xy/(x+y)^2+yz/(y+z)^2+zx/(z+x)^2
而x/(y+z)+y/(z+x)+z/(x+y)-3/2
=[(x-y)^2/(y+z)(z+x)+(y-z)^2/(x+y)(z+x)
...全部
我们知道sin^2A=1/(1+cot^2A)
令x=cotA,y=cotB,z=cotC,可以得到xy+yz+zx=1,
原不等式化为
1/(x+y)^2+1/(y+z)^2+1/(z+x)^2≥9/4 (*)
先看一个变形
1/(x+y)^2=(xy+yz+zx)/(x+y)^2=xy/(x+y)^2+z/(x+y)
所以(*)右边
=x/(y+z)+y/(z+x)+z/(x+y)+xy/(x+y)^2+yz/(y+z)^2+zx/(z+x)^2
而x/(y+z)+y/(z+x)+z/(x+y)-3/2
=[(x-y)^2/(y+z)(z+x)+(y-z)^2/(x+y)(z+x)
+(z-x)^2/(x+y)(y+z)]/2
xy/(x+y)^2+yz/(y+z)^2+zx/(z+x)^2-3/4
=[(x-y)^2/(x+y)^2+(y-z)^2/(y+z)^2+(z-x)^2/(z+x)^2]/4
至此(*)等价于
2[(x-y)^2/(y+z)(z+x)+(y-z)^2/(x+y)(z+x)
+(z-x)^2/(x+y)(y+z)]
≥(x-y)^2/(x+y)^2+(y-z)^2/(y+z)^2+(z-x)^2/(z+x)^2
在x≥y≥z时，上式左边一三项不小于右边一三项，但第二项不一定
所以我们只需证明
2(z-x)^2/(z+x)^2-(z-x)^2/(z+x)^2
≥(y-z)^2/(y+z)^2-2(y-z)^2/(x+y)(z+x)
去分母得
(x-z)^2[2(x+z)^2-(x+y)(y+z)](y+z)
≥(y-z)^2[(x+y)(x+z)-2(y+z)^2](x+z)
而x-z≥y-z,且(x-z)(y+z)-(y-z)(z+z)=2x(x-y)≥0
所以
(x-z)^2[2(x+z)^2-(x+y)(y+z)](y+z)
-(y-z)^2[(x+y)(x+z)-2(y+z)^2](x+z)
≥(y-z)^2(x+z)[2(z+x)^2-(x+y)(y+z)-(x+y)(x+z)-2(y+z)^2]
=(y-z)^2(x+z)[(x-y)^2+2xz+2yz+4z^2]
≥0
至此，原不等式获证！。
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