Question

Show that for all real numbers, this statement is greater than or equal to 0.

Answer 1

\(\LARGE\color{black}{ \rm{this~~statement} \ge0 }\) like this?

Answer 2

This statement.\[\LARGE a^2+b^2+c^2 - ab-bc-ac \ge 0\]

Answer 3

can you turn it into a function of x and then see if there are any roots between x = 0 and 1?

Answer 4

oh that wont do it it has to also approach infinity as x approaches infinity

Answer 5

and that there are no roots from 0 to infinity

Answer 6

I just discovered a method to solve this today, I wonder if there's an alternate way though. ;)

Answer 7

does your solution involve limits of infinity?

Answer 8

Nope

Answer 9

I guess I'll share my answer, since it's been a while. Stop me if you want to try though.

Answer 10

Take the total derivative and notice that the derivative is zero when \(a = b = c\).

Answer 11

If \(a = b =c\) then \(a^2+b^2+c^2−ab−bc−ac = 0\)

Answer 12

Therefore the minimum is zero.

Answer 13

\[a^2+b^2+c^2-ab-ac-bc \] Looking at the last three terms, and allowing \[\omega = \frac{-1}{2}+i \frac{\sqrt{3}}{2}\] We can split each of these into two parts, one half in the positive i and one half in the negative i direction, which are just complex conjugates. \[a^2+b^2+c^2+ \omega (ab+ac+bc)+ \omega^2 (ab+ac+bc) \] We can now rewrite this as \[(a+ \omega b + \omega^2c)(a + \omega^2b+ \omega c)\] which is really just (a+wb+w^2c) multiplied by its complex conjugate which is the square modulus of a complex number which is always positive. ;P

Answer 14

\(\displaystyle\frac{\partial \left(x^2-x y-x z+y^2-y z+z^2\right)}{\partial \{x,y,z\}} = \{2 x-y-z,-x+2 y-z,-x-y+2 z\}\)

Answer 15

Solution to {2 x - y - z, -x + 2 y - z, -x - y + 2 z} = {0, 0, 0}
\(y = x, z = x\)

Answer 16

So, you can solve it using analysis as a minimization problem.

Answer 17

I guess I could, but I'd rather not after figuring out his method haha.

Answer 18

I think it is more practical.

Answer 19

I am not disagreeing with you, it's nice to have a "brute force" method of doing things, but sometimes it's nice to find something that works as well.

For instance, my method has its own way of being generalized and looked at that's slightly different and gives a different look at the situation.

Answer 20

I found another nice proof

WLOG we can assume \(a\le b\le c\) and appeal to below inequality :
http://en.wikipedia.org/wiki/Rearrangement_inequality

Answer 21

Oh interesting, I am not sure I understand this argument though @ganeshie8 can you explain it?

Answer 22

suppose \(a_1\le a_2\le \cdots \le a_n\) and \(b_1\le b_2\le \cdots \le b_n\)

then
\(a_1b_n + a_2b_{n-1}+\cdots + a_nb_{1} ~~\le~~ a_1b_1+a_2b_2+\cdots + a_nb_n\)

Answer 23

It has more to that inequality. If we consider all the sums of products above both sums give the least and upper bounds

Answer 24

wiki has more clear explanation and proof..

Answer 25

present proof is a one liner if we use that inequality :
suppose
\[a\le b\le c\]
\[a\le b\le c\]

below is the max of all product of sums by the inequality in that link: \[aa + bb + cc \]

Answer 26

we use the (without lose of generality law )
which means assume some order of the three values of a,b,c

Answer 27

hehe seems ganesh already post same thing