Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers 
Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers

It is in the exercises of the AM-GM inequality chapter of a book, and that is why I believe it will be solved by that. Can anyone give me a proof using that or otherwise, too?
 A: We use AM-GM three times
$$\frac{a^2 +a^2+a^2+a^2 +b^2+c^2}{6} \geq a (abc)^{1/3}$$
$$\frac{b^2 +b^2+b^2+b^2 +a^2+c^2}{6} \geq b (abc)^{1/3}$$
$$\frac{c^2 +c^2+c^2+c^2 +b^2+a^2}{6} \geq c (abc)^{1/3}$$
Summing these inequalities and dividing by 6 gives 
$$a^2+b^2+c^2 \geq (a+b+c)(abc)^{1/3}$$
Now using that $abc=1$, we conclude the desired inequality.
A: The previous answer was not a full answer, let me offer my thoughts and let me expand a little (using Cauchy-Schwartz):
Using the Cauchy-Schwarz inequality we get
\begin{align}
a+b+c=a\cdot 1+b\cdot 1+c\cdot 1+\leq\sqrt{a^2+b^2+c^2}\sqrt{1^2+1^2+1^2}\tag{1}
\end{align}
From the AM-GM inequality we obtain $a^2+b^2+c^2\geq 3\sqrt[3]{a^2b^2c^2}=3$, so
\begin{align}
\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\tag{2}
\end{align}
From $(1)$ and $(2)$ it clearly follows that
\begin{align}
a+b+c\leq\sqrt{a^2+b^2+c^2}\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}=a^2+b^2+c^2
\end{align}
Further Edit using only AM-GM
Reduce the problem to elementary algebra by considering
$$a^2 \ge 2a -1 \tag3$$
$$b^2 \ge 2b-1 \tag4$$
$$c^2 \ge 2c-1 \tag5$$
Then add up $(3), (4)$ & $(5)$ and get:
$$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \tag6$$
By AM-GM we have 
$$\frac{a+b+c}{3} \ge (abc)^\frac{1}{3}=1 $$
$$ a+b+c \ge 3 \tag7 $$
Finally, from $(6)$ and $(7)$ we obtain the required inequality:
$$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \ge a+b+c $$
A: We can also proceed as follows:
\begin{align*}
a^2+b^2+c^2 &\ge \frac{(a+b+c)^2}{3}\\
&= (a+b+c)\,\frac{a+b+c}{3}\\
&\ge (a+b+c) \sqrt[3]{abc}\\
&=a+b+c.
\end{align*}
A: To add to the otherwise category, here is a hint:
It is enough to show $f(x)=x^2-x-\log x$ is never negative for any positive $x$, which is easy from univariate calculus.
