prove

[abc]

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

[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

[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.

[Gokul43201]

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

[s0l0m0nsh0rt]

**Sigh**

I miss the dinosaurs. -ss
:rofl:

[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.

[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.

[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

[Gokul43201]

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

[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