Problem 10.2. Find all values of the real parameter $a$ such that the inequality

$$
\sqrt{4+3 x} \geq x+a
$$

has no an integer solution.

Stoyan Atanassov
10.2. It is enough to find all reals $a$ such that the converse inequality

$$
\sqrt{4+3 x}2$ is a necessary condition.

We shall prove that it is also a sufficient condition. Indeed, for $a>2$ we have $x+a>x+2$ and it is enough to prove that $x+2 \geq \sqrt{4+3 x}$ for every integer $x \geq-1$. We consecutively have

$$
x+2 \geq \sqrt{4+3 x} \Leftrightarrow\left|\begin{array}{l}
(x+2)^{2} \geq 4+3 x \\
x \geq-\frac{4}{3}
\end{array} \Leftrightarrow\right| \begin{aligned}
& x^{2}+x \geq 0 \\
& x \geq-\frac{4}{3}
\end{aligned}
$$

The later system of inequalities is satisfied for $x \in\left[-\frac{4}{3},-1\right] \cup[0, \infty)$, i.e. for every integer $x \geq-1$. Therefore the solutions of the problem are $a \in(2, \infty)$.