Prove a^2+b^2+c^2

1. Aug 5, 2004

abc

can any one prove this :
a^2+b^2+c^2 ( is greater or equal to ) ab + ac + bc
thanx
regards
abc

2. Aug 5, 2004

arildno

A very important identity in mathematics is:
1+1=2
Note that this is readily transformed into another important identity:
$$\frac{1}{2}+\frac{1}{2}=1$$
now, try to rewrite
$$(\frac{1}{2}+\frac{1}{2})a^{2}+(\frac{1}{2}+\frac{1}{2})b^{2}+(\frac{1}{2}+\frac{1}{2})c^{2}-ab-ac-bc$$

3. Aug 5, 2004

s0l0m0nsh0rt

This is like....
a^2+b^2 >= ab, right?

take (a-b)^2 >= 0.
(a-b)^2=a^2+b^2-2ab.
so a^2+b^2 >= 2ab >=ab.

so now u want a^2+b^2+c^2,
so take (a-b-c)^2, and (a+b-c)^2, etc...
and do the same sort of thing, a bit trickier though.

4. Aug 5, 2004

Gokul43201

Staff Emeritus
arildno's approach is much nicer...but forgive his efforts at humor.

5. Aug 5, 2004

s0l0m0nsh0rt

**Sigh**

I miss the dinosaurs. -ss
:rofl:

6. Aug 5, 2004

HungryChemist

How about...we assume a^2 + b^2 + c^2 < ab + ac + bc and consider a special case where 'a' being minus and absolute value of 'a' is greater than the absolute value of b for ovious reason to lead a contradiction. So it will prove the negation of what assumed is true.

7. Aug 6, 2004

matt grime

it will only provide a contradiction in the special case where a is negative and in abs value greater then b, which need not be true.

8. Aug 7, 2004

maverick280857

Use the identity for $$a^2 + b^2 + c^2 - ab - bc - ca$$:

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

The right hand side is always greater than or equal to zero (equality in the case a = b = c). This proves the result.

Hope that helps...

Cheers
Vivek

Last edited: Aug 7, 2004
9. Aug 7, 2004

Gokul43201

Staff Emeritus
This is exactly what arildno was saying - without actually putting the spoon in the mouth.

10. Aug 7, 2004

maverick280857

Oh well I didn't quite figure that out and since the question seemed unanswered to me so I went ahead and posted the solution (put the spoon in the mouth if you like it that way) :-). Its been quite a while since it was posted anyway.

Cheers
Vivek