3. Determine all values of the real parameter $a$ such that for every $x \in[-1,1]$ the inequality

$$
a x^{2}+2(a+1) x+a-4 \leq 0
$$

holds.
Solution. We transform the given inequality into powers of $x+1$ and obtain

$$
a(x+1)^{2}+2(x+1)-6 \leq 0
$$

We introduce the substitution $x+1=t$ and obtain the inequality

$$
a t^{2}+2 t-6 \leq 0
$$

which must be satisfied for every $t \in[0,2]$.

For $t=0$ we have $a \cdot 0^{2}+2 \cdot 0-6=-6 \leq 0$ and this holds for every $a \in \mathbb{R}$.

For $t \in(0,2]$ we have

$$
a \leq \frac{6-2 t}{t^{2}}=2\left(\frac{3}{t^{2}}-\frac{1}{t}\right)=2\left(3 u^{2}-u\right), \text { where } u=\frac{1}{t} \in\left[\frac{1}{2},+\infty\right)
$$

The inequality $a \leq 2\left(3 u^{2}-u\right)$ must hold for every $u \in\left[\frac{1}{2},+\infty\right)$, so it is sufficient for the parameter $a$ to be bounded from above by the minimum of the quadratic function $2\left(3 u^{2}-u\right)$ on the given interval, i.e., it must hold that $a \leq \min _{\left.u \in \frac{1}{2},+\infty\right)} 2\left(3 u^{2}-u\right)$. The quadratic function $f(u)=2\left(3 u^{2}-u\right)$ has a minimum at the vertex $u=\frac{1}{6}$ and increases for $u>\frac{1}{6}$. Then for $u \in\left[\frac{1}{2},+\infty\right)$, the minimum is reached at $u=\frac{1}{2}$. Therefore, it must hold that $a \leq 2\left(3\left(\frac{1}{2}\right)^{2}-\frac{1}{2}\right)=\frac{1}{2}$, which means $a \in\left(-\infty, \frac{1}{2}\right]$.