又一个非钝角不等式

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！。
  收起