danlu
幼苗
共回答了23个问题采纳率:91.3% 举报
设△ABC面积为s,三条边分别为a b,c
则s=1/2*a *ha =1/2*b*hb=1/2*c*hc =rp (p=1/2(a+b+c))
所以ha=2s/a hb=2s/b hc=2s/c r=2s/(a+b+c)
代入ha+hb+hc=9r得 2s/a+2s/b+2s/c=9*2s/(a+b+c)
得1/a+1/b+1/c=9/(a+b+c)
得(a+b+c)(1/a+1/b+1/c)=9
a+b+c>=3(abc)^(1/3) 1/a+1/b+1/c>=3(1/(abc))^(1/3)
所以(a+b+c)(1/a+1/b+1/c)>=9
显然只有a=b=c时等号才成立.
1年前
2