$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$ 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 $a^2 + b^2 + c^2 + 6 \ge 3(a+b+c)$.
 A: 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.
A: 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.
A: 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 $x > \frac{1}{2}$.
A: 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.
