<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] As A.M. of any set of positive number ≥ G.M,
$\frac{a+b+c}{3}≥ (abc)^{\frac{1}{3}}$
Now, $  3(a^2+b^2+c^2)-(a+b+c)^2= (a-b)^2+(b-c)^2+(c-a)^2≥ 0$
=>$ a^2+b^2+c^2≥ \frac{(a+b+c)^2}{3}$
=>$ a^2+b^2+c^2≥(a+b+c)(\frac{a+b+c}{3})≥a+b+c$  as $\frac{a+b+c}{3}≥1$
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