How to prove: $a+b+c\le a^2+b^2+c^2$, if $abc=1$? Let $a,b,c \in \mathbb{R}$, and $abc=1$. What is the simple(st) way to prove inequality 
$$
a+b+c \le a^2+b^2+c^2.
$$
(Of course, it can be generalized to $n$ variables).

By replacing $a, b, c$ by $|a|, |b|, |c|$ if needed, we may assume they are non-negative. Then just apply Jensen inequality (or AM-GM inequality) to deduce that
$$ a+b+c = \sum_{\text{cyclic}} a^{4/3}b^{1/3}c^{1/3} \leq \sum_{\text{cyclic}} \frac{4}{6}a^{2} + \frac{1}{6}b^{2} + \frac{1}{6}c^{2} = a^2 + b^2 + c^2. $$
