$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$ I have not solved inequalities in a while, so I am a little rusty. Could you help me with this inequality I have found?
$$a^2 + b^2 + c^2 + 6 \ge 3(a + b + c),$$ where $a, b, c > 0$ and $abc = 1$
My initial idea was $a ^ 2 + 2 \ge 2\sqrt 2a$ and the inequalities with $b $ and $c$, then adding these 3, we get, $a ^ 2 + b ^ 2 + c ^ 2 + 6 \ge 2\sqrt 2(a + b+ c)$, but then we get to $2\sqrt2 > 3$, which is false.
Edited: I found some variants of the original problem.
Problem 1: Let $a, b, c > 0$. Prove that $a^2 + b^2 + c^2 + 6 + (abc - 1) \ge 3(a+b+c)$.
Problem 2: Let $a, b, c$ be reals with $abc \le 1$. Prove that $a^2 + b^2 + c^2 + 6 \ge 3(a+b+c)$.

Let $t = \sqrt{ab}$ and
$$f(a,b,c) = a^2+b^2+c^2 + 6 - 3(a+b+c).$$
Suppose $c = \min\{a,\,b,\,c\},$ then $c \leqslant 1,$ so
$$\sqrt{a}+\sqrt{b} \geqslant 2\sqrt{t} \geqslant 2.$$
We have
$$f(a,b,c) - f(t,t,c) = a^2+b^2-2t^2 - 3(a+b-2t) = \left[\left(\sqrt{a}+\sqrt{b}\right)^2-3\right]\left(\sqrt{a} - \sqrt{b}\right)^2 \geqslant 0.$$
Therefore $f(a,b,c) \geqslant f(t,t,c),$ and
$$f(t,t,c) = f\left(t,t,\frac{1}{t^2}\right) = \frac{(2t^4-2t^3+2t+1)(t-1)^2}{t^4} \geqslant 0.$$
The proof is completed.
