an inequality?

Question

If a,b,c are postive real numbers and abc=1 .

Show that 1+ 3/(a+b+c)≧6/(ab+bc+ca)

aakash v , i don't understand your answer

Answer 1

this can be proved by conventional method..
in LHS
1+3/(a+b+c)=abc+3/(a+b+c)
abc(a+b+c)+3/a+b+c
a^2bc+b^2ac+c^2ab+3/(a+b+c)
in abc=1 we have a=1/bc b=1/ac and c=1/ab
substuting in RHS we get
6/(a/c+b/a+c/b)
6/(a^2bc+b^2ac+c^2ab/abc)
6/(a^2bc+b^2ac+c^2ab) which is obviosly less than LHS and equal to LHS when all are 1