Question

a^2+b^@+c^2=1 then bc+ca+ab lies in interval??

Answer 1

b^???

Answer 2

2

Answer 3

@dumbcow , @experimentX , @dpaInc

Answer 4

@Mani_Jha

Answer 5

Use QM>= AM >= GM

Answer 6

hw??? @Aron_West

Answer 7

(a+b+c)^2=a2+b2+c2+2(bc+ab+ca)
(a+b+c)^2=1+2(bc+ab+ca)
(bc+ab+ca)=0.5[(a+b+c)^2 - 1]=0.5(a+b+c)^2 -0.5
then ,what to do????

Answer 8

Quadratic Mean >= Arithmetic Mean >= Geometric mean

Answer 9

Also use Harmonic Mean.

Answer 10

where did you get this q from @aravindG

Answer 11

must be less than 3 and greater than -3 .. thats for sure.

Answer 12

how @experimentX???

Answer 13

a,b,c must be less than 1 and greater than 1

Answer 14

ok...

Answer 15

but the problem states that what would be the least upper bound and greatest lower bound.

Answer 16

....i m not getting dat

Answer 17

you know the answer???

Answer 18

no

Answer 19

@Sarkar it came in my test!!

Answer 20

Oh .. i found answer on google

Answer 21

you could use the method of lagrange multipliers

Answer 22

if you still want to solve then beware of google!!!

Answer 23

It lies between -1/2 and 1.Simple!

Answer 24

\[(a+b+c)^{2}=a ^{2}+b ^{2}+c ^{2}+2(ab+bc+ca)\]
Let\[ab+bc+ca=x\]
Thus RHS becomes 1+2x
AM>=GM
\[(1+2x)/2\ge \sqrt{2x}\]
Now square, factorize and obtain a range. Thanks to @Aron_West and @Sarkar for the hints.

Answer 25

cool @mani...

Answer 26

The only pre-requites are ":
Any square of a real no.>0 & AM>= GM

Answer 27

thx all

Answer 28

@Mani_Jha nice one :D

Answer 29

To get the bounds :
(a+b+c)^2 >=0 or,1+2x>=0 or x>=-1/2
Now use AM>=GM

Answer 30

Thanks @anonymoustwo44, and @Aron_West, yes even that condition is to be considered.