How can the following inequation be proven?
$a^2 + b^2 + c^2 \ge ab + bc + ca$
How can the following inequation be proven?
$a^2 + b^2 + c^2 \ge ab + bc + ca$
Try $(a-b)^2+(b-c)^2+(c-a)^2 \ge0$
Compute lhs, divide by two and rearrange.
This is a specific form of Cauchy-Schwarz inequality.
Let $x = (a, b, c)$ and $y = (b, c, a)$ as vectors.
The inequality is $ | \left< x,y \right>| \le \|x\|\|y\|. $ with standard inner product definition. One neat trick to prove this is using an auxilary parameter $t,$ and expanding $ \| x+ty \|^2 = \left< x+ty,x+ty \right> = \|x\|^2 + 2 \left< x,y \right>t +\|y\|^2t^2.$ We know, this being a square, is non-negative. Therefore, the discriminant of the polynomial in $t$ is less or equal to zero. Which is $\left< x,y \right>^2 - (\|x\|\|y\|)^2 \le 0.$ Substituting the values for $x$ and $y$ will do the job.
$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0$
From Cauchy-Schwarz
$ab+bc+ac=\sqrt{a^2}\sqrt{b^2}+\sqrt{b^2}\sqrt{c^2}+\sqrt{a^2}\sqrt{c^2} \leq \sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}$
Moving on;
$ab+bc+ac \leq a^2+b^2+c^2$
Done!
İf $\ c> a ,a^2+c^2 \gt 2ac $because$\ (a-c)^2 \gt 0$ İf $\ c>b>a ,c^2+b^2/2+a^2/2 \gt ac+bc $ and $\ b^2/2+a^2/2 \gt ab $ sum of them $\ a^2+b^2+c^2 \gt ab+bc+ac $ İf $\ c=b \gt a $ or $\ a=b=c $ it can be solved with same logic.