How to prove: $a+b+c\le a^2+b^2+c^2$, if $abc=1$?

Question

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).

Answer 1

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!

Answer 2

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 $$ a^2+b^2+c^2 \ge ~|a|+|b|+|c| ~>~ a+b+c. $$

Answer 3

We may assume that $a+b+c > 0$.

Then you can do this by using Cauchy-Schwarz and the power mean inequality as follows:

$$a+b+c \le \sqrt{a^2 + b^2 + c^2} \sqrt{3} \le \sqrt{a^2 + b^2 + c^2} \sqrt{a^2 + b^2 + c^2} = a^2 + b^2 + c^2$$

Answer 4

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 $\prod a_i = 1$.

Answer 5

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

$$ a+b+c = \sum_{\text{cyclic}} a^{4/3}b^{1/3}c^{1/3} \leq \sum_{\text{cyclic}} \frac{4}{6}a^{2} + \frac{1}{6}b^{2} + \frac{1}{6}c^{2} = a^2 + b^2 + c^2. $$