10.1. Find the sum of all integer values of $a$ belonging to the interval $[-10 ; 10]$, for each of which the double inequality $5 \leq x \leq 10$ implies the inequality $a x+3 a^{2}-12 a+12>a^{2} \sqrt{x-1}$.
Answer: -47. Solution. Let $t=\sqrt{x-1}$, then the original problem reduces to finding such values of $a$ for which the inequality $f(t) \equiv a t^{2}-a^{2} t+3 a^{2}-11 a+12>0$ holds for all $t \in[2 ; 3]$. This means that the minimum of the function $f(t)$ on the interval $2 \leq t \leq 3$ is positive.

If $a=0$, then $f(t) \equiv 12$. If $a \neq 0$, then there are two cases:
a) $\left\{\begin{array}{l}f\left(\frac{a}{2}\right)=-\frac{1}{4}(a-2)(a-4)(a-6)>0, \\ 2 \leq \frac{a}{2} \leq 3\end{array} \Leftrightarrow 40, \\ f(3)=2(6-a)>0\end{array} \Leftrightarrow a<3\right.$.

In the end, $a \in(-\infty ; 3) \cup(4 ; 6)$, and the sum of the sought values of $a$ on the interval $[-10 ; 10]$ is: $-10-9-8-\ldots-1+0+1+2+5=(-10-9-\ldots-6)-4-3=-47$.