2014/089) show that if $abc = 1$ then $a^2+b^2+c^2\ge a + b + c$

we have by AM GM inaquality

$a^2+1 \ge\ 2a \cdots (1) $
$b^2+1 \ge\ 2b \cdots (2) $
$c^2+1 \ge\ 2c \cdots (3) $

further

$\dfrac{a+b+c}{3} \ge \sqrt[3]{ abc}$
or  $\dfrac{a+b+c}{3} \ge 1$
or $a+b+c \ge 3$
or $(a+b+c-3 )\ge 0\cdots(4) $

adding  (1) (2) and (3) we get

$a^2+b^2+c^2 + 3 \ge 2(a+b+c)$
or $a^2+b^2+c^2  \ge a+b+c + (a+b+c-3)$
using (4) we get
 $a^2+b^2+c^2  \ge a+b+c$