if a,b ,c >0 and (a+b+c)(ab+ab+ca)=3z value of z is

a) zabc
c)z=abc
rest options i forgot.please suggest the ans

• formula -----> (a+b+c)(ab+bc+ca)>=9abc
  
  therefore 9abc=3z
  
  so z=3abc
  
  So, 3abc is the answer.
• 9 years agoHelpfull: Yes(3) No(2)
• Cosider a=b=c=1 as a,b,c>0.Substitute in the given eqution we get z=3
  
• 9 years agoHelpfull: Yes(1) No(0)
• Formula is (a+b+c)(ab+bc+ca)>=9abc
  here we have (a+b+c)(ab+bc+ca)=3z
  so, 3z>=9abc
  z>=3abc
  so ans is z is greater than 3abc
  
• 8 years agoHelpfull: Yes(1) No(0)
• a+b+c)(ab+ab+ca)=3z , z>3abc that's the answer.....(a+b+c)(an+BC+can)= 9abc
• 9 years agoHelpfull: Yes(0) No(1)
• it gives out of range warning
• 9 years agoHelpfull: Yes(0) No(1)
• (a+b+c)(ab+bc+ca)=3z
  take the value of a=b=c=1/3
  pt in this qust.
  1(1/9+1/9+1/9)=3z
  1/3=3z
  z=1/9
  so,
  ans=(a+b+c)/3
  
• 9 years agoHelpfull: Yes(0) No(3)