if a,b,c are real and (bc + ca + ab)^2 k (a+b+c) , then k =?

Question

if a,b,c are real and  (bc + ca + ab)^2 > k (a+b+c) , then k =?

anonymous · Answer

@satellite73   @experimentX    @ujjwal  @UnkleRhaukus @Ishaan94   @dumbcow  @shubhamsrg

experimentx · Answer

k=0

anonymous · Answer

lol....$k=0$  workin

shubhamsrg · Answer

LHS = b^2c^2 + c^2a^2 + a^2b^2 + 2(a^2bc + ab^2c + abc^2) 
       =b^2c^2 + c^2a^2 + a^2b^2 + 2abc(a+b+c) 
hence,
(ab + bc +ca)^2 = (some positive no.) + 2abc(a+b+c)
thus (ab+bc+ca)^2 > 2abc(a+b+c)

then i guess k =2abc ??

note that  (ab+bc+ca)^2 = 2abc(a+b+c) when a=b=c

anonymous · Answer

The answer should be 3

anonymous · Answer

@shubhamsrg

shubhamsrg · Answer

ohh..lol..

shubhamsrg · Answer

leme try and see..

anonymous · Answer

ok

shubhamsrg · Answer

i dont see 3 as an ans..
for a=0,b=1,c=0 ,,3 fails!!

shubhamsrg · Answer

2abc doesnt seem to fail for me ,,,

shubhamsrg · Answer

if we see it as AM>GM ,
(a+b)^2 >4ab =>a^2 + b^2 >2ab
similarily,
b^2 + c^2 > 2bc
c^2 + a^2 > 2ca
adding all,
a^2 + b^2 +c^2 > ab+bc+ca
a^2 + b^2 + c^2 + 2(ab+bc+ca) > 2(ab+bc+ca) + ab+bc+ca
(a+b+c)^2 > 3(ab+bc+ca)

lol..it turned the other way around this time!! :D

shubhamsrg · Answer

its symmetrically different from what the ques asks..

anonymous · Answer

@Yahoo! 

plz check the question again..........lool

experimentx · Answer

well ... i would say the same!! if k = 3