3. Find all triples of real numbers $a, b, c$, for which the set of solutions to the inequality

$$
\sqrt{2 x^{2}+a x+b}>x-c
$$

with the unknown $x$ is the set $(-\infty, 0\rangle \cup(1, \infty)$.

(P. Černek)
Solution. Let $(*)$ denote the given inequality and $K=(-\infty, 0) \cup(1, \infty)$ the corresponding set (of all) solutions. From the fact that 0 belongs to $\mathrm{K}$, it follows for $b$ the condition $b \geqq 0$ and simultaneously $\sqrt{b}>-c$. However, if $b>0$ were to hold, the expression $\sqrt{2 x^{2}+a x+b}$ would be defined in some neighborhood of the point $x=0$ and from (the strict) inequality $(*)$ for $x=0$, its validity would also follow for small positive numbers $x$, which is in contradiction with the form of the set $K$. Therefore, $b=0$ must hold, and from the inequality $\sqrt{b}>-c$ the condition $c>0$ follows.

Since $\sqrt{2 x^{2}+a x+b}=\sqrt{x(2 x+a)}$ and since the set $K$ contains all numbers $x>1$, the inequality $2 x+a \geqq 0$ holds for such $x$, which means that $a \geqq-2$. Since $1 \notin \mathrm{K}$, the inequality $\sqrt{2+a}>1-c$ does not hold, but its left side has meaning due to the inequality $a \geqq-2$. Therefore, on the contrary, $\sqrt{2+a} \leqq 1-c$ holds, from which the condition $c \leqq 1$ follows. If the strict inequality $\sqrt{2+a}0$, which is in contradiction with the fact that $1+\varepsilon \in \mathrm{K}$. This means that $\sqrt{2+a}=1-c$, from which $a=(1-c)^{2}-2=c^{2}-2 c-1$.

Summarizing the results of our considerations: we have found that every suitable triplet of numbers $(a, b, c)$ is necessarily of the form $\left(c^{2}-2 c-1,0, c\right)$, where $0x-c
$$

for a fixed $c \in(0,1\rangle$ and corresponding $a=c^{2}-2 c-1$.

From the inequalities $00$. Substituting $a=c^{2}-2 c-1$ yields the inequality $(x-1)\left(x+c^{2}\right)>0$, which holds for precisely those (positive) numbers $x \in\left\langle-\frac{1}{2} a, \infty\right)$ that are greater than 1 (reiterating that $-\frac{1}{2} a \leqq 1$). This proves that the set of solutions of the inequality (1) is indeed the set $\mathrm{K}$ from the problem statement.

Answer: The sought triplets are $(a, b, c)=\left(c^{2}-2 c-1,0, c\right)$, where $c$ is any number from the interval $(0,1\rangle$.