Question

An interesting problem , I came across that i would like to share.

Answer 1

If a ,b ,c > 0
(b+c-a)(c+a-b)(a+b-c) -abc is

(a)Positve
(b)Negative
(c)Non-positive
(d)Non-negative

Answer 2

AM-GM inequality gives a nice proof :)

Answer 3

So which option

Answer 4

c

Answer 5

It is b actually

Answer 6

ahh yes ! it is 0 only when a,b,c are all 0

Answer 7

you want to prove how you got b ? :)

Answer 8

Yes! , no i just wanted to share it , i know how to get to b
actually :)
How do you know if a,b,c are all not zero

Answer 9

if a,b,c are all 0, then the expression equals 0 and it wont be negative anymore !

Answer 10

share your proof also if possible :)

Answer 11

but you now know it is negative, actually not my proof , i saw the sollution after thinking on it

Answer 12

yes, anything is fine

Answer 13

You are also right

Answer 14

i think the key thing to notice is :

\[\large (a+b-c) + (b+c-a)+(c+a-b) = a+b+c\]

Answer 15

then by AM-GM inequality :

\[\large \sqrt[3]{(a+b-c) (b+c-a)(c+a-b)} \lt \dfrac{a+b+c}{3}\]

Answer 16

If there is one of a,b,c, which is greater than the other two, exactly one of the trinomial factors is negative, so the first product is negative and since abc must be positive, the result must be negative.

Answer 17

Yes both are correct

Answer 18

Consider a , b , c as sides of a triangle

Answer 19

The sum of two sides is always greater than the third side

Answer 20

So , (b+c-a)(c+a-b)(a+b-c) is always positive

Answer 21

Applying AM-GM inequality , we arrive at the result that they are negative

Answer 22

applying AM-GM inequality is the tricky part, how do u apply ? i still feel stuck :o

Answer 23

looks below won't yield anything useful

\[\large \sqrt[3]{(a+b-c) (b+c-a)(c+a-b)} \lt \dfrac{a+b+c}{3}\]

Answer 24

if two of a,b,c are equal, the product of the trinomials is a negative number times a square, which must be negative, so the result is negative.

Answer 25

\[\frac{(b+c-a)+(c+a-b) }{ 2 } \ge \sqrt{(b+c-a)(c+a-b)}\]

Answer 26

yes the problem with that logic is it wont work when a,b,c are sides of triangle @LarsEighner

Answer 27

nice xD

Answer 28

Taking more such AM-GM inequalities with different pairs
WE will get abc on L.H.S when we multiply all three eqautions

Answer 29

It came in 1986 nice problem

Answer 30

got it completely :) somehow i have been wasting time messing around three terms lol

Answer 31

However it appears to me (I could be wrong) if a=b=c, then the expression evaluates to zero. So it is not in general negative.

Answer 32

yeah thats the reason they constrained a,b,c > 0

Answer 33

if we let a,b,c to be negative, the expression gives positives also i guess

Answer 34

need to check...

Answer 35

So the answer appears to be d.

Answer 36

It is b actually

Answer 37

If a=b=c, a,b,c > 0

b+c-a)(c+a-b)(a+b-c) -abc =>
(2a -a)^3 - a^3
(a)^3 - a^3 = 0

b is clearly wrong and d is the answer.

Answer 38

I am not trying to argue with you sir , but it was an official exam called IIT , which had this question and the answer was "B"

Answer 39

The exam was wrong. Algebra does not lie. Test makers sometimes make mistakes.

Answer 40

that looks interesting

a = b = c

a^3 - a^3 = 0

hmm so the question needs to be reviewed

Answer 41

that should make \(c\) as the answer right ? non-positive

Answer 42

Oh yes. It is c.