<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 second solution:
WLOG, assume $c = \min(a,b,c)$.
If $c \le \frac{1}{2}$, we have
$$a^2+b^2+c^2 + 6 - 3(a+b+c) = (a-\tfrac{3}{2})^2 + (b - \tfrac{3}{2})^2 + c^2 - 3c + \tfrac{3}{2}
\ge 0.$$
If $c > \frac{1}{2}$, noting that $x^2 + 2 - 3x + \ln x \ge 0$ for all $x > \frac{1}{2}$
(see the remark at the end),
we have
$$a^2+b^2+c^2 + 6 - 3(a+b+c) = \sum_{\mathrm{cyc}} (a^2 + 2 - 3a + \ln a) \ge 0.$$
We are done.
Remark: Let $f(x) = x^2 + 2 - 3x + \ln x$. We have $f'(x) = \frac{(2x-1)(x-1)}{x}$.
Thus, $f(x)$ is strictly decreasing on $(\frac{1}{2}, 1)$, and strictly increasing on $(1, \infty)$.
Also $f(1) = 0$. Thus, we have $f(x) \ge 0$ for all $x > \frac{1}{2}$.
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