Question

A cubic polynomial F has three "different" real zeros a,b and c. The coefficient of x^3 is positive. Show that F`(a)+F`(b)+F`(c) > 0

Answer 1

\[F(x)=p(x-a)(x-b)(x-c)\] is probably a good start, where \(p>0\)
take the derivative via the product rule and see what you get
i didn't actually do it, but is seems like a start to me

Answer 2

Thanks, would that be all I need to do?

Answer 3

i don't know
i am writing it now
but it seems like the only way to start

Answer 4

\[F'(x)=p(x-b)(x-c)+p(x-a)(x-b)+p(x-a)(x-c)\] is a start
then compute \(F'(a)\) etc. add. and see what you get

Answer 5

Not sure how to do that either, I've only just begun attempting this type of problems.

Answer 6

All help is very appreciated

Answer 7

man this is not nearly as easy as i though

Answer 8

lets ignore the \(p\) for a second, as it is positive, lets say it is 1

Answer 9

\[F'(x)=(x-b)(x-c)+(x-a)(x-b)+(x-a)(x-c)\]
\[F'(a)=(a-b)(a-c)\]
\[F'(b)=(b-a)(b-c)\]
\[F'(c)=(c-a)(c-b)\]

Answer 10

add them up and get
\[(a-b)(a-c)+(b-a)(b-c)+(c-a)(c-b)\]

Answer 11

there might be some elementary reason why this is positive, but i don't see it
if you do a raft of algebra you can write this as the sum of two squares, and so it has to be positive (non negative actually)

Answer 12

oh actually positive, since it is only zero if \(a=b\) or \(b=c\) and you are told that they are different

Answer 13

@Zarkon do you see right of the bat that this must be positive?

Answer 14

look at \[\frac{1}{2}[(a-b)^2+(a-c)^2+(b-c)^2]\]

Answer 15

*off

Answer 16

\[f'(a)+f'(b)+f'(c)=\frac{1}{2}[(a-b)^2+(a-c)^2+(b-c)^2]\]

Answer 17

my f being your F ;)

Answer 18

suppressing the p

Answer 19

i believe you but i have to figure out why

Answer 20

Same here

Answer 21

oh, algebra is why

Answer 22

Heh, could you explain it? I feel as though I am not smart enough to figure it out by myself at the moment :/

Answer 23

it is algebra:
multiply out, combine like terms, and see what you get
although how @Zarkon recognized it as
\[\frac{1}{2}\left((a-b)^2+(a-c)^2+(b-c)^2\right)\] is for him to know

Answer 24

Does that prove that F`(a)+F`(b)+F`(c) > 0 is possible?

Answer 25

a square is always positive unless it is zero

Answer 26

it was a guess

Answer 27

really? damn! good guess

Answer 28

I thought it might be that ...so I expanded both

Answer 29

@Dahlioz multiply it out and see that it works

Answer 30

i cheated and used wolfram, probably would have been better to do it by hand, which i did afterwards

Answer 31

wolf gives this
http://www.wolframalpha.com/input/?i=%28a-b%29%28a-c%29%2B%28b-a%29%28b-c%29%2B%28c-a%29%28c-b%29
although not
\[\frac{1}{4}\left(2a-b-c\right)^2+\frac{3}{4}\left(b-c\right)^2\] which also gives it

Answer 32

ah...cool

Answer 33

not nearly as cool as the sum of three squares

Answer 34

I agree ;)

Answer 35

Thanks guys, you both have been very helpful

Answer 36

Just one question, about what grade/year of school would this be suitable for?

Answer 37

But why can't (2a-b-c) be equal to 0?

Answer 38

As wolfram gives 1/4 (2 a-b-c)^2+3/4 (b^2-2 b c+c^2) rather than 1/4(2a−b−c)^2+3/4(b−c)^2. @Zarkon Sorry for disturbing you guys with this.

Answer 39

those two expressions are equal

Answer 40

Ahh, I see