<s>[INST] $a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$ 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 $a^2 + b^2 + c^2 + 6 \ge 3(a+b+c)$.
 [/INST] By using Schur's inequality and the identity
\begin{align}
&(a+b+c)^3 + 9abc - 4(a+b+c)(ab + bc + ca)\\
 =\ & a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b),
\end{align}
we have $(a+b+c)^3 + 9abc - 4(a+b+c)(ab + bc + ca) \ge 0$ which results in
$$\frac{(a+b+c)^3 + 9abc }{4(a+b+c)} \ge ab + bc + ca. \tag{1}$$
We need to prove that $(a+b+c)^2 - 2(ab + bc + ca) + 6 \ge 3(a+b+c)$.
By using (1), it suffices to prove that
$$(a+b+c)^2 - 2\cdot \frac{(a+b+c)^3 + 9abc }{4(a+b+c)} + 6 \ge 3(a+b+c)$$
that is (using $abc = 1$)
$$\frac{(a+b+c - 3)[(a+b+c)^2 - 3(a+b+c) + 3]}{2(a+b+c)} \ge 0$$
which is true (using $a+b + c \ge 3\sqrt[3]{abc} = 3$).
We are done.
Remark: Actually, it is the pqr method.
</s>