a分之1 +2b分之1 +3c分之1=1,求证a+2b+3c≥9

需加条件a,b,c>0
∵1/a+1/(2b)+1/(3c)=1
∴a+2b+3c
=(a+2b+3c)[1/a+1/(2b)+1/(3c)]
=3+a/(2b)+a/(3c)+2b/a+2b/(3c)+3c/a+3c/(2b)
=3+[a/(2b)+2b/a]+[a/(3c)+3c/a]+[2b/(3c)+3c/(2b)]
≥3+2√[a/(2b)*2b/a]+2√[a/(3c)*3c/a]+2√[2b/(3c)*3c/(2b)]
=3+2+2+2=9
∴a+2b+3c≥9