if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$ This is supposed to be an application of AM-GM inequality.

if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$

First of all,
$a^2+b^2+c^2\ge 3$
by a direct application of AM-GM.Also,we have
$a^2+b^2+c^2\ge ab+bc+ca$
Next,we consider the expression
$(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc\le a+b+c+a^2+b^2+c^2+1$
but that hardly helps.I know that
$3(a^2+b^2+c^2)\ge (a+b+c)^2$
From the first derived inequality,we know that the left hand side of the above inequality is greater than or equal to 9.But I can't see how that can be used here.I know that we can get $a+b+c\ge 3$ using AM-GM but that does not take me a step closer to finding the solution(from what I can understand).Also,
$a^3+b^3+c^3\ge a^2b+b^2c+c^2a$
$(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\ge a^2b+b^2c+c^2a$
A hint will be appreciated at this point.
 A: Use the fact that 
$$a^2+a^2+a^2+b+c \geq5\sqrt[5]{a^2.a^2.a^2.b.c}=5\sqrt[5]{a^6bc}=5a$$ since $abc=1$
Similarly, we get two more results, and adding them we get:
$$3(a^2+b^2+c^2)+2(a+b+c) \geq 5(a+b+c) $$ which gives us our result.
$$a^2+b^2+c^2 \geq a+b+c$$
$$Q.E.D.$$
A: Here is an exotic solution based on geometry.
Let $\mathcal{M}$ and $\mathcal{S}$ be surfaces defined by
\begin{align*}
\mathcal{M} : abc = 1
\quad \text{and} \quad
\mathcal{S} : a^{2} + b^{2} + c^{2} = a + b + c.
\end{align*}
Then we have the following observations:


*

*$\mathcal{M}$ lies outside the sphere of radius $\sqrt{3}$. Indeed, if $X = (a, b, c) \in \mathcal{M}$, then the square-distance from the origin $O$ satisfies
$$ \overline{OX}^{2} = a^{2} + b^{2} + c^{2} \geq 3\sqrt[3]{a^{2}b^{2}c^{2}} = 3. $$

*$\mathcal{S}$ is contained in the sphere of radius $\sqrt{3}$. Indeed, $\mathcal{S}$ is the sphere of radius $\frac{\sqrt{3}}{2}$ centered at the point $P =  (\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$, hence if $X = (a, b, c) \in \mathcal{S}$ then by the triangle inequality
$$ \overline{OX} \leq \overline{OP} + \overline{PX} = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}.$$
Combining two facts, we find that $\mathcal{M}$ lies outside of $\mathcal{S}$ (possible except for the tangent point). Therefore
\begin{align*}
abc = 1
& \quad \Longrightarrow \quad (a, b, c) \in \mathcal{M} \\
& \quad \Longrightarrow \quad (a, b, c) \text{ lies outside } \mathcal{S} \\
& \quad \Longrightarrow \quad \left(a - \tfrac{1}{2}\right)^{2} + \left(b - \tfrac{1}{2}\right)^{2} + \left(c - \tfrac{1}{2}\right)^{2} \geq \tfrac{3}{4} \\
& \quad \Longrightarrow \quad a^{2} + b^{2} + c^{2} \geq a + b + c.
\end{align*}
A: Alternatively, $a^2+a^2+a^2+a^2+b^2+c^2 \geq 6 \sqrt[6]{a^8b^2c^2} = 6 \sqrt[6]{a^6} = 6|a| \geq 6a$ by AM-GM. Adding the analogous inequalities $a^2+4b^2+c^2 \geq 6b$ and $a^2+b^2+4c^2 \geq 6c$ gives the result.
A: Since $(2,0,0)\succ\left(\frac{4}{3},\frac{1}{3},\frac{1}{3}\right)$, our inequality it's just Muirhead.
A: Notice that for all real $a,b,c$,
$$(a - 1)^2 + (b-1)^2 + (c - 1)^2 \ge 0.$$
$$a^2 + b^2 + c^2 - 2a - 2b - 2c + 3 \ge 0.$$
$$a^2 + b^2 + c^2 \ge -3 + (a + b + c) + (a + b + c).$$
But by AM-GM, $a + b + c \ge 3\sqrt[3]{abc} = 3$. So,
$$a^2 + b^2 + c^2 \ge -3 + 3 + (a + b + c).$$
$$a^2 + b^2 + c^2 \ge  a + b + c.$$
Equality is attained when $a = b = c = 1$.
(I really wish to put everything after the first line as a spoiler so that this answer becomes a hint, but I don't know how D:)
A: $GM\le AM\le RMS\implies GM\times AM\le RMS^2\implies a+b+c\le a^2+b^2+c^2$.
