Proof inequality

**Ideasman** · June 22nd 2007, 09:52 AM

For all real numbers a, b, c, prove that:

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

Thanks!

**ThePerfectHacker** · June 22nd 2007, 10:35 AM

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,
$a^2 - ab + b^2 - bc + c^2 - ac \geq 0$
Iff,
$a^2+b^2+c^2 \geq ab+bc+ac$
Note, by Cauchy-Swartz
$a^2+b^2+c^2 \geq |ab+bc+ac|$
But,
$|ab+bc+ac|\geq ab+bc+ac$
So,
$a^2+b^2+c^2 \geq ab+bc+ac$
And the inequality is true.

**Ideasman** · June 22nd 2007, 01:53 PM

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|

**Jhevon** · June 22nd 2007, 01:58 PM

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

**ThePerfectHacker** · June 22nd 2007, 02:36 PM

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

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

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

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

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

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