Question

已知a>0,b>0,c>0,a+b+c=1,求证1/a+1/b+1/c≥9

Answer 1

已知a>0,b>0,c>0,a+b+c=1,求证1/a+1/b+1/c≥9
(a+b+c)(1/a+1/b+1/c)
=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c
=1+(b+c)/a+1+(a+c)/b+1+(a+b)/c
=3+(a/b+b/a)+(a/c+c/a)+(b/c+c/b)
≥3+2+2+2=9

1/a+1/b+1/c≥9
祝你好运
吉林 汪清LLX

Answer 2

a+b+c=1
那么有:
(1/a+1/b+1/c)
=(a+b+c)(1/a+1/b+1/c)
=3+a/b+b/a+a/c+c/a+b/c+c/b,
利用基本不等式，a/b+b/a>=2,
a/c+c/a>=2,b/c+c/b>=2,
三式相加
所以3+a/b+b/a+a/c+c/a+b/c+c/b>=9