五月

终于鼓求出来了。感谢豪神的引理(三次多项式判别式)

引理如下

①
1-a=x,1-b=y,1-1/(ab)=z
a=1-x,b=1-y,ab=1/(1-z)=(1-x)(1-y)
因此(1-x)(1-y)(1-z)=1
xyz-(xy+yz+zx)+(x+y+z)=0
1-(1/x+1/y+1/z)+1/(xy)+1/(yz)+1/(zx)=0
1/x^2+1/y^2+1/z^2=(1/x+1/y+1/z)^2-2[1/(xy)+1/(yz)+1/(zx)]=(1/x+1/y+1/z-1)^2+1≥1