For all real numbers a, b, c, prove that:
a(a - b) + b(b - c) + c(c - a) >= 0.
Thanks!
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.
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,
$\displaystyle a^2+b^2+c^2 \geq |ab+bc+ac|$