Proof inequality

• Jun 22nd 2007, 09:52 AM
Ideasman
Proof inequality
For all real numbers a, b, c, prove that:

a(a - b) + b(b - c) + c(c - a) >= 0.

Thanks!
• Jun 22nd 2007, 10:35 AM
ThePerfectHacker
Quote:

Originally Posted by Ideasman
For all real numbers a, b, c, prove that:

a(a - b) + b(b - c) + c(c - a) >= 0.

Open paranthesis,
$\displaystyle a^2 - ab + b^2 - bc + c^2 - ac \geq 0$
Iff,
$\displaystyle a^2+b^2+c^2 \geq ab+bc+ac$
Note, by Cauchy-Swartz
$\displaystyle a^2+b^2+c^2 \geq |ab+bc+ac|$
But,
$\displaystyle |ab+bc+ac|\geq ab+bc+ac$
So,
$\displaystyle a^2+b^2+c^2 \geq ab+bc+ac$
And the inequality is true.
• Jun 22nd 2007, 01:53 PM
Ideasman
You'd have to prove the cauchy-shwartz thm wouldn't you?

I don't see how

a^2 + b^2 + c^2 >= |ab + bc + ac|
• Jun 22nd 2007, 01:58 PM
Jhevon
Quote:

Originally Posted by Ideasman
You'd have to prove the cauchy-shwartz thm wouldn't you?

I don't see how

a^2 + b^2 + c^2 >= |ab + bc + ac|

It's a well-known theorem, so i don't suppose you would have to prove it. However, if it makes you feel better, you shouldn't have much trouble finding a proof on the internet, or maybe TPH would like to give you one
• Jun 22nd 2007, 02:36 PM
ThePerfectHacker
Quote:

Originally Posted by Ideasman
You'd have to prove the cauchy-shwartz thm wouldn't you?

I don't see how

a^2 + b^2 + c^2 >= |ab + bc + ac|

We know that,
$\displaystyle \ \left(\sum_{k=1}^n a_k^2\right) \cdot \left( \sum_{k=1}^n b_k^2 \right) \geq \left( \sum_{k=1}^n a_kb_k \right)^2$

Let, $\displaystyle a_1=a , \ a_2 = b , \ a_3 = c$
And, $\displaystyle b_1 = b, \ b_2 = c, \ b_3 = a$
With $\displaystyle n=3$

Then,
$\displaystyle (a^2+b^2+c^2)^2=(a^2+b^2+c^2)(b^2+c^2+a^2) \geq (ab+bc+ac)^2$

Take square roots, and note that $\displaystyle \sqrt{(ab+bc+ac)^2} = |ab+bc+ac|$

So,
$\displaystyle a^2+b^2+c^2 \geq |ab+bc+ac|$