Prove that $a^{2}+b^{2}+c^{2}-a b-b c-c a$ is always non-negative for all values of $a$, $b$ and $c$.
We have,
$a^{2}+b^{2}+c^{2}-a b-b c-c a$
Multiply and divide by ‘2’
$=2 / 2\left[a^{2}+b^{2}+c^{2}-a b-b c-c a\right]$
$=1 / 2\left[2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a\right]$
$=1 / 2\left[a^{2}+a^{2}+b^{2}+b^{2}+c^{2}+c^{2}-2 a b-2 b c-2 c a\right]$
$=1 / 2\left[\left(a^{2}+b^{2}-2 a b\right)+\left(a^{2}+c^{2}-2 c a\right)+\left(b^{2}+c^{2}-2 b c\right)\right]$
$=1 / 2\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]\left[?(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$=\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2} \geq 0$
$\therefore a^{2}+b^{2}+c^{2}-a b-b c-c a \geq 0$
Hence, $a^{2}+b^{2}+c^{2}-a b-b c-c a \geq 0$ is always non-negative for all values of $a, b$ and $c$.
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