# How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$?

How can the following inequation be proven?

$$a^2 + b^2 + c^2 \ge ab + bc + ca$$

• "How can it be proved" - not solved. There is nothing to solve here . – davidlowryduda Sep 15 '11 at 20:00
• I think $a,b,c$ should be greater or equal with $0$. – Iuli Sep 6 '17 at 11:28

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.

This is also a consequence of the Rearrangement inequality.

• beautiful mathematics – LoveFood Mar 11 '14 at 20:08

$$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0$$

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.