设a.b.c为三角形的三边 求证:a/(b+c-a) b/(a+c-b) c/(a+b-c)≥3

设a.b.c为三角形的三边 求证:a/(b+c-a)+b/(a+c-b) +c/(a+b-c) ≥3（？）
证明：设b+c-a=x，a+c-b=y，a+b-c=z
因为a.b.c为三角形的三边，故：x, y, z＞0
故：a+b+c=x+y+z
故：a=(y+z)/2，b=(x+z)/2，c=(x+y)/2
故：a/(b+c-a)+b/(a+c-b) +c/(a+b-c)
=(y+z)/(2x)+ (x+z)/(2y)+ (x+y)/(2z)
=1/2(y/x+x/y)+1/2(x/z+z/x)+1/2(y/z+z/y)
≥1/2×2+1/2×2+1/2×2=3
故：a/(b+c-a)+b/(a+c-b) +c/(a+b-c) ≥3

a/(b +c-a)+ b/(a +c-b) +c/(a+ b-c)=
=2a/2(b +c-a)+ 2b/2(a +c-b) +2c/2(a+ b-c)=
=(a+b+c)/2(b +c-a)-1/2+ (a+b+c)/2(a +c-b)-1/2 +(a+b+c)/2(a+ b-c)-1/2=
=1/2(a+b+c)[1/(b +c-a)+ 1/(a +c-b)+1/(a+ b-c)]-3/2=
=1/2[(b +c-a)+ (a +c-b)+(a+ b-c)][1/(b +c-a)+ 1/(a +c-b)+1/(a+ b-c)]-3/2
≥(1/2)*3^2-3/2=3.