Proving $a^2 + b^2 + c^2 \geqslant ab + bc + ca$

Question

Let $a, b, c \in \mathbb{R}$. Show that $a^2 + b^2 + c^2 \geqslant ab + bc + ca$.

My reasoning went as follows and I would like to know if it's correct.

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

$\Leftrightarrow a^2 + b^2 + c^2 -ab - bc - ca \geqslant 0$

$\Leftrightarrow 2a^2 + 2b^2 + 2c^2 -2ab - 2bc - 2ca \geqslant 0$

$\Leftrightarrow a^2 + a^2 + b^2 + b^2 + c^2 + c^2 -2ab - 2bc - 2ca \geqslant 0$

$\Leftrightarrow (a^2 -2ab + b^2) + (b^2 - 2bc + c^2)+ (c^2 - 2ca + a^2) \geqslant 0$

$\Leftrightarrow (a -b)^2 + (b-c)^2 + (c-a)^2 \geqslant 0$.

And the last inequality holds since squares are non-negative.

I've seen couple of different ways of doing this and they seem to divide by $2$ and I didn't really understand it so instead can I just use the fact that $2a^2 = a^2+a^2$?

Answer 1

Your proof is correct and so is the identity $2a^2=a^2+a^2$. Alternatively, if you want to divide by $2$, then you can write

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

$\Leftrightarrow\frac {a^2 + b^2}{2} + \frac {b^2 + c^2}{2} + \frac {a^2 + c^2}{2}\geqslant ab + bc + ca$

$\Leftrightarrow\frac {a^2 + b^2}{2} + \frac {b^2 + c^2}{2} + \frac {a^2 + c^2}{2}-ab- bc - ca\geqslant 0$

$\Leftrightarrow\frac {a^2 -2ab +b^2}{2} + \frac {b^2 -2bc +c^2}{2} + \frac {a^2 -2ac +c^2}{2}\geqslant 0$

$\Leftrightarrow(a-b)^2 + (b-c)^2 + (c-a)^2 \geqslant 0$

Answer 2

Add up the inequalities below

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

to arrive at,

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

Answer 3

Yes, you can.

But I think it's better to release this way and the fact $2a^2=a^2+a^2$ by the cyclic sum: $$a^2+b^2+c^2-ab-ac-bc=\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=$$ $$=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0.$$ Another way: $$a^2+b^2+c^2-ab-ac-bc=a^2-(b+c)a+b^2-bc+c^2=$$ $$=\left(a-\frac{b+c}{2}\right)^2+b^2-bc+c^2-\frac{(b+c)^2}{4}=\left(a-\frac{b+c}{2}\right)^2+\frac{3}{4}(b-c)^2\geq0.$$