Inequality $a+b+c > abc$, prove $a^2+b^2+c^2 > abc$ Given, $$a+b+c > abc$$
Prove $$a^2+b^2+c^2 > abc$$
I tried to square the first one but it wouldn't work.

$$a^2+b^2+c^2\geq ab+bc+ac$$ holds,  since it is equivalent with $$(a-b)^2+(b-c)^2+(c-a)^2\geq 0.$$ Also $$ab+bc+ac\geq\sqrt{3abc(a+b+c)}.$$ Previous inequality holds, since it is equivalent with $$a^2b^2+b^2c^2+c^2a^2\geq abc(a+b+c),$$
which is equivalent with $$(x-y)^2+(y-z)^2+(z-x)^2\geq 0$$ for $x=bc$, $y=ac$, $z=ab$.
Finally, $$a^2+b^2+c^2\geq ab+bc+ac\geq\sqrt{3abc(a+b+c)}\geq\sqrt{3}abc>abc$$ 
