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

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

Thanks!

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- Jun 22nd 2007, 09:52 AMIdeasmanProof 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 AMThePerfectHacker
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 PMIdeasman
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 PMJhevon
- Jun 22nd 2007, 02:36 PMThePerfectHacker
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,

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