Question

If a,b,c are positive real number such that b+c-a,c+a-b and a+b-c are positive the expression
(b+c-a)(c+a-b)(a+b-c) is
A).positive
B).negative
C).non-positive
D).nonnegative

Answer 1

Oh wait it's wrong question..

Answer 2

-abc is also there in the question

Answer 3

where will "-abc" come?
please edit your question

Answer 4

(b+c-a)(c+a-b)(a+b-c)-abc

Answer 5

we can simply assume appropriate values for a,b and c
and then get the answer..

Answer 6

I want a proper ans though..

Answer 7

No approximation pls

Answer 8

thats not approximation because there is only 1 solution to this question which must hold true for all values of a,b,c which follow those 3 equations..

you can do this too-assume a,b,c to be side lengths of a triangle

Answer 9

by 3 equations i mean
a+b-c>0
b+c-a>0
c+a-b>0
triangle inequalities

Answer 10

Bt afterwards we have to Assume some value of a,b,c which satisfy these three equation..

Answer 11

@ilovepuppieslol do u know of another way??

Answer 12

what are your thoughts on this question? :|

Answer 13

Lol...

Answer 14

...

Answer 15

I opened up the brackets ..and came up with a very big expression

Answer 16

\(a^3 +b^3+c^3-3abc=(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac) \)

\(\Large \frac{a^3 +b^3+c^3-(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac) }{3}=abc\)



\((b+c-a)(c+a-b)(a+b-c)-abc\)

\((b+c-a)(c+a-b)(a+b-c)-\Large\frac{a^3 +b^3+c^3-(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac) }{3}\)

Answer 17

Did u do it like that?

Answer 18

no
i just shared another form of the equation

Answer 19

Oh .i see

Answer 20

Well.
I feel now that your first approach to the problem was nice...
For if i solve it like this it wud be very time consuming...

Answer 21

yeah
a typical way of doing this would involve expanding the stuff and that would take time :(

Answer 22

Thanks

Answer 23

np