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).

We may assume that $a+b+c > 0$.
Then you can do this by using Cauchy-Schwarz and the power mean inequality as follows:
$$a+b+c \le \sqrt{a^2 + b^2 + c^2} \sqrt{3} \le \sqrt{a^2 + b^2 + c^2} \sqrt{a^2 + b^2 + c^2} = a^2 + b^2 + c^2$$
