TITLE: How to prove: $a+b+c\le a^2+b^2+c^2$, if $abc=1$?
QUESTION [9 upvotes]: 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).

REPLY [5 votes]: A).
If values $~a,b,c~$ are positive, then
denote $m = \dfrac{a+b+c}{3},~$  ($m\ge \sqrt[3]{abc}=1$, AM-GM);    
$\left\{\begin{array}{l}
a=m+\alpha,\\ 
b=m+\beta, ~~~~~~(\alpha+\beta+\gamma=0).\\ 
c=m+\gamma.
\end{array}\right.~~~
$
Then 
$$
a^2+b^2+c^2 ~=~ 3m^2 + \alpha^2+\beta^2+\gamma^2 ~\ge~ 3m ~=~ a+b+c.
$$
B). If some (two) of values $~a,b,c~$ are negative, then
$$
a^2+b^2+c^2 \ge ~|a|+|b|+|c| ~>~ a+b+c.
$$
{"set_name": "stack_exchange", "score": 9, "question_id": 501106}