
Problem 10.1. Find all values of the real parameter $a$ such that the inequality $x^{4}+a x^{3}+(a+3) x^{2}+a x+1>0$ holds for all real values of $x$.


Solution: If $x=0$, the inequality holds for any $a$. Suppose $x \neq 0$. Dividing both sides of the inequality by $x^{2}$ gives $x^{2}+a x+a+3+$ $\frac{a}{x}+\frac{1}{x^{2}}>0$ or $\left(x+\frac{1}{x}\right)^{2}+a\left(x+\frac{1}{x}\right)+a+1>0$. Substitute $t=x+\frac{1}{x}$. If $x>0$, it is true that $t \geq 2$, and if $x0$ holds for any $t \in(-\infty ;-2] \cup[2 ;+\infty)$. Denote $f(t)=t^{2}+a t+a+1$. The discriminant of $f(t)$ is $D=a^{2}-4 a-4$.

Let $D0$ holds for any real $t$, in particular for $t \in(-\infty ;-2] \cup[2 ;+\infty)$.

Let $D \geq 0$, i.e., $a \in(-\infty ; 2-\sqrt{2}] \cup[2+\sqrt{2} ;+\infty)$. Then the inequality $f(t)>0$ holds for any $t \in(-\infty ;-2] \cup[2 ;+\infty)$ if and
only if

$$
\left\lvert\, \begin{aligned}
& -20 \\
& f(-2)>0
\end{aligned}\right.
$$

The solutions of this system when $a \in(-\infty ; 2-\sqrt{2}] \cup[2+\sqrt{2} ;+\infty)$ are

$$
a \in\left(-\frac{5}{3} ; 2-2 \sqrt{2}\right] . \text { So } a \in\left(-\frac{5}{3} ; 2+2 \sqrt{2}\right)
$$
