Question

Prove that a^2+b^2+c^2-ab-bc-ca is always non-negative for all values of a, b and c.

Answer 1

1/2 ((a-b)^2 +(b-c)^2+(c-a)^2)

Answer 2

ok this is an important elementary inequality :

start with

\[(a-b)^2+(b-c)^2+(a-c)^2 \ge 0\]

Answer 3

\[(a-b)^2+(b-c)^2+(a-c)^2 \ge 0\]\[2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\]\[a^2+b^2+c^2-ab-bc-ca\ge0\]

Answer 4

Thanks for the solutions guys, though I am familiar with this solution. However, I was looking for something beginning from (a+b+c)^2 or something like that....Is it possible to solve it any other way???

Answer 5

It's a fairly well known identity, I suppose you could start by just doubling the initial expression to make the factorization slightly more clear (this sort of cyclic solution is very common in inequalities).
(There might be a way to do it as you suggest, let me think about it, might be a problem to categorically demonstrate that all the bits are positive).

Answer 6

If we take the Law of Cosines (so only works for positive a,b,c):

a^2 = ...., b^2 = ...., c^2 =.... and add the three equations up -->

(a^2+b^2 +c^2) = a^2 + b^2 + c^2 - (ab Cos + etc) which works because Cos is just a scaling factor..

Still thinking...

Answer 7

a^2 + b^2 +c^2 >= ab + bc + ca is a direct consequence of the so-called Rearrangement equality (as are many other identities) but I am guessing you are looking for something more elementary...

Answer 8

anyway thanks guys for suggesting the solution that u hv already given....