<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] My third solution:
Actually $a^2 + b^2 + c^2 + 6 - 3(a+b+c) + (abc - 1) \ge 0$ for all $a, b, c \ge 0$.
(pqr method)
Let $p = a + b + c$, $q = ab + bc + ca$ and $r = abc$.
By Schur's inequality $a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) \ge 0$
which is written as $p^3 - 4pq + 9r \ge 0$, we have $\frac{p^3 + 9r}{4p}\ge q$.
We need to prove that $p^2 - 2q + 6 - 3p + r - 1 \ge 0$.
It suffices to prove that
$p^2 - 2 \cdot \frac{p^3 + 9r}{4p} + 6 - 3p + r - 1 \ge 0$ or
$$\frac{(2p-9)r}{2p} + \frac{1}{2}p^2 - 3p + 5 \ge 0. \tag{1}$$
If $2p- 9 \ge 0$, clearly (1) is true.
If $2p - 9 < 0$, since $\frac{p^3}{27} \ge r$, it suffices to prove that
$$\frac{(2p-9)\frac{p^3}{27}}{2p} + \frac{1}{2}p^2 - 3p + 5 \ge 0$$
that is
$$\frac{1}{27}(p+15)(p-3)^2 \ge 0.$$
We are done.
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