$a^2+b^2+c^2 \geq 3 abc$ I have to solve this Prove that
 $a^2+b^2+c^2 \geq 3 abc$ 
given that
 $a+b+c \geq 3 abc$
I am stuck!


If $a,b,c\in \Bbb R$, and if $a+b+c\ge 3abc$, then: $$a^2+b^2+c^2\ge 3abc$$

Proof:

*

*Case 1:
If $abc\le 1$, then we use AM-GM inequality, and have:
$$a^2+b^2+c^2=|a|^2+|b|^2+|c|^2\ge 3\sqrt[3]{|abc|^2}\ge 3abc$$
this is true because
$$\Longleftrightarrow |abc|^2\ge (abc)^3$$


*Case 2:
If $abc\ge 1$, since: $$a+b+c\ge 3abc>0$$
we can use Cauchy-Schwarz inequality in the following way:
$$a^2+b^2+c^2\ge\dfrac{1}{3}(a+b+c)^2\ge 3a^2b^2c^2=3(|abc|)^2\ge 3abc$$
