已知a，b，c∈R+，a+b+c=1，求证：1a+1b+1c≥9．

Question

已知a，b，c∈R₊，a+b+c=1，求证：(1)/(a)+(1)/(b)+(1)/(c)≥9．

Answer 1

证明：由题意知(1)/(a)+(1)/(b)+(1)/(c)=(a+b+c)/(a)+(a+b+c)/(b)+(a+b+c)/(c)
=3+（(b)/(a)+(a)/(b)）+（(c)/(a)+(a)/(c)）+（(b)/(c)+(c)/(b)）
∴(b)/(a)+(a)/(b)≥2，(c)/(a)+(a)/(c)≥2，(b)/(c)+(c)/(b)≥2．
当且仅当a=b=c时，取等号，
∴(1)/(a)+(1)/(b)+(1)/(c)≥9．