Question

a^2+b^2+c^2+ab+bc+ac>0 for all a, b, c?
please, help

Answer 1

I need prove or disprove the expression

Answer 2

You can try relating it to (a+b+c)^2 by adding and subtracting terms.

Answer 3

I did, but go nowhere (a+b+c)^2 -ab-bc-ac and then?

Answer 4

Try this then

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

Answer 5

We know that this is always true. Expand and reach your inequality.

Answer 6

\(\large a^2 + b^2 + c^2 +ab+bc+ca = \frac{1}{2}[(a+b)^2 +(b+c)^2 + (c+a)^2 ]\)

Answer 7

which is > 0, if \(a\ne -b, b\ne -c, c\ne -a\)

Answer 8

u can compress it into a single constraint

Answer 9

Thank you. I know what the result is now. So, it is not >0 for all. counter example is a = 1, b= -1 c =-1?

Answer 10

nope

Answer 11

try a = 1, b = -1, c = 1

Answer 12

ahh doesnt work either

Answer 13

if it >0 , need prove, if it is not >0, just counter example. That's all
@AravindG I expand your expression, it goes to something different from original one :)

Answer 14

your counter example doesnt seem to work... check once

Answer 15

yes, it doesn't hehehe

Answer 16

counter example : a = 0, b = 0, c = 0

Answer 17

not an exciting one.. .

Answer 18

I misread the question and tried to prove a^2 + b^2 + c^2 ≥ ab + ac + bc. Sorry.

Answer 19

but if a = b=0, c =-1 works

Answer 20

doesnt work

Answer 21

again, you are right, me wrong, hehehehe

Answer 22

If we just use the expansion kindly provided by @ganeshie8, and the fact that it's equal to 0 if and only if \(a=-b, b= -c, c=-a\), then we can solve the expression in terms of \(a\). We get\[b=-a\]\[c=-b=a\]but also\[c=-a\]So \(c=a=-a\). This is only true if \(a=0\), and consequently \(a=b=c=0\). So other than the trivial solution where they're all 0, your expression must always be greater than 0.

Answer 23

wow ! that looks really cool+simple proof xD

Answer 24

Thank you very much, I got it.