- 一道不等式设a,b,c为三角形的三边，求证：a/b+c
- 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.
- 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）-（b+c-a)]/2(b+c-a)+[（a+b+c）-（a+c-b)]/2(a+c-b)+[（a+b+c）-（a+b-c)]/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
=(a+b+c)/2 [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×{√[(b+c-a)×1/(b+c-a)] + √[(a+c-b)×1/(a+c-b)] + √[(a+b-c)×1/(a+b-ac)]}^2 -3/2
=(1/2)×(1+1+1)^2-3/2
=3
得a/b+c-a +b/a+c-b +c/a+b-c≥3