<s>[INST] How prove $abc+1>3a$ if $a^2+b^2+c^2=9,0Let $a,b,c\in\mathbb{R}$ and such $0<a\le b\le c$, and $ a^2+b^2+c^2=9$. Show that
$$abc+1>3a$$
Since $$9=a^2+b^2+c^2\ge 3a^2\Longrightarrow a^2\ge 3$$
then I can't, maybe can use AM-GM inequality to solve it?
 [/INST] From $b \le c$ and $a^2+b^2+c^2= 9$, we have
$$b^2c^2 \ge a^2c^2 = a^2(9-a^2-b^2) \ge a^2(9-2a^2)$$
Now $2a^2 \le 6 < 9 \implies 9-2a^2 > 0$, so the above gives us $bc \ge a\sqrt{9-2a^2}$.  As $a$ is positive, $\implies abc \ge a^2\sqrt{9-2a^2}$.
Thus it is enough to show for $a \in (0, \sqrt3]$, we must have
$$a^2\sqrt{9-2a^2} \ge 3a-1 \iff f(a) = 2a^6-9a^4-(3a-1)^2 \le 0 \tag{1}$$
Case 1:  $0 < a \le 1$
We note that $f$ can be written as $f(a) = 2a^4(a^2-1)-7a^4-(3a-1)^2 < 0$
Case 2:  $1 < a \le \sqrt{3 \over 2}$
Note $f(a) = \left(a^4+\frac32 \right)(2a^2-3)-6a^2(a^2-1)-6\left(a-\frac{11}{12}\right)< 0 $
Case 3:  $\sqrt{3 \over 2} < a \le \sqrt{3}$
Note $f(a)=a^2(a^2-3)(2a^2-3)+(1-6a) < 0 $
Thus in all cases, we find $abc+1 > 3a$.

P.S. The inequality $abc+1 \ge 3a$ would hold for all real $a, b, c$ s.t. $a \le b \le c$ and $a^2+b^2+c^2=9$, positivity is needed only for the strict inequality version.
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