均值不等式已知a>0b>0c>0且a+b+c=1求证1/a+1/b+1/c≥9

Question

均值不等式
已知a>0 b>0 c>0 且 a+b+c=1 求证 1/a + 1/b +1/c ≥9

Answer 1

因为a+b+c=1,a>0 b>0 c>0
故：1/a + 1/b +1/c
=(a+b+c)/a + (a+b+c)/b +(a+b+c)/c
=1+b/a+c/a++a/b+1+c/b+a/c+b/c+1
=3+(a/b+b/a)+(a/c+c/a)+(b/c+c/b)
≥3+2+2+2=9
当a/b=b/a,a/c=c/a,b/c=c/b,即：a=b=c时,取等号