<s>[INST] Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that 
$a^2+b^2+c^2 \geq a+b+c$. 
Thanks
 [/INST] Let's solve it in an elementary way and start from the fact that:
$$a^2 \ge 2a -1 \tag1$$
$$b^2 \ge 2b-1 \tag2$$
$$c^2 \ge 2c-1 \tag3$$
Then add up $(1)$ $(2)$ $(3)$ and get:
$$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \tag4$$
By AM-GM we have 
$$\frac{a+b+c}{3} \ge (abc)^\frac{1}{3}=1 $$
$$ a+b+c \ge 3 \tag5 $$
Finally, from $(4)$ and $(5)$ we obtain the required inequality:
$$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \ge a+b+c $$
Q.E.D.
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