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).
 A: Assume $a,b,c>0$.
$$\text{Cauchy Schwarz:}\;\;3(a^2+b^2+c^2)\geq \left(a+b+c\right)^2\qquad\text{AM-GM}:a+b+c\geq 3$$
Done!
A: 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
$$
a^2+b^2+c^2 \ge ~|a|+|b|+|c| ~>~ a+b+c.
$$
A: We may assume that $a+b+c > 0$.
Then you can do this by using Cauchy-Schwarz and the power mean inequality as follows:
$$a+b+c \le \sqrt{a^2 + b^2 + c^2} \sqrt{3} \le \sqrt{a^2 + b^2 + c^2} \sqrt{a^2 + b^2 + c^2} = a^2 + b^2 + c^2$$
A: Another way, we note that it is sufficient to prove for positive variables.  Now let $f(x) = x^2 - x - \log x$.  Then the  inequality is equivalent to $f(a) + f(b) + f(c) \ge 0$  
It is easy to see $f$ has a minimum at $x = 1$ when $f(1) = 0$, so $f(x) \ge 0$.  
By this approach it is quite easy to extend to $\sum f(a_i) \ge 0$, given $\prod a_i = 1$.
A: 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
$$ a+b+c = \sum_{\text{cyclic}} a^{4/3}b^{1/3}c^{1/3} \leq \sum_{\text{cyclic}} \frac{4}{6}a^{2} + \frac{1}{6}b^{2} + \frac{1}{6}c^{2} = a^2 + b^2 + c^2. $$
