Let's determine all real numbers $x$ that satisfy the inequality

$$
\sqrt{3-x}-\sqrt{x+1}>\frac{1}{2}
$$
I. solution. The left side is real if and only if there is no negative number under any of the square root signs:

$$
x+1 \geqq 0 \quad \text { and } \quad 3-x \geqq 0, \quad \text { thus } \quad-1 \leqq x \leqq 3 \text {. }
$$

The inequality

$$
\sqrt{3-x}>\sqrt{x+1}+\frac{1}{2}
$$

when written in the form under (2), neither side is negative, so this inequality holds if and only if (1) also holds. Therefore, for $x$ between -1 and 3, we need to solve the inequality obtained by squaring and rearranging:

$$
\frac{7}{4}-2 x>\sqrt{x+1}
$$

Here, the left side cannot be negative. This imposes another restriction on $x$:

$$
x<\frac{7}{8} \quad \text { and } \quad x \geqq -1
$$

Squaring both sides again, we get:

$$
\left(\frac{7}{4}-2 x\right)^2>x+1
$$

Expanding and rearranging, we have:

$$
4 x^2 - 12 x + \frac{33}{16} > 0
$$

Multiplying through by 16 to clear the fraction:

$$
64 x^2 - 192 x + 33 > 0
$$

This can be factored as:

$$
4\left(x-\frac{8+\sqrt{31}}{8}\right)\left(x-\frac{8-\sqrt{31}}{8}\right)>0
$$

This inequality holds if and only if both differences are positive or both are negative:

$$
x>1+\frac{\sqrt{31}}{8}, \quad \text { or } x<1-\frac{\sqrt{31}}{8}
$$

The first contradicts (4), while the second provides a stronger restriction than (4). Combining this with (2), the values of $x$ we seek are:

$$
-1 \leq x<1-\sqrt{31} / 8
$$

Hegedús Csaba (Nagykanizsa, Landler J. g. III. o. t.)

![](https://cdn.mathpix.com/cropped/2024_05_02_374b08013105c1d2000fg-1.jpg?height=572&width=634&top_left_y=1541&top_left_x=735)

II. solution. The function $\sqrt{3-x}-\sqrt{x+1}$ is meaningful in the interval $-1 \leqq x \leqq 3$. The function decreases as $x$ increases. Indeed, the square root of a larger number is larger, and that of a smaller number is smaller, so if $x$ increases, $3-x$ and thus $\sqrt{3-x}$ decrease, while $x+1$ and $\sqrt{x+1}$ increase, and therefore $-\sqrt{x+1}$ decreases. Consequently,

$$
\sqrt{3-x}-\sqrt{x+1}=\sqrt{3-x}+(-\sqrt{x+1})
$$

the function, being the sum of two decreasing terms, also decreases.

At $x = -1$, the function value is 2, which is greater than 1/2, so we can answer the problem by finding the number $x_{0}$ for which the function takes the value $1 / 2$. Then (1) holds for the values where $-1 \leqq x<x_{0}$. The

$$
\sqrt{3-x}-\sqrt{x+1}=\frac{1}{2}
$$

equation, after squaring twice and rearranging, yields:

$$
\begin{gathered}
4-2 \sqrt{(3-x)(x+1)}=\frac{1}{4}, \quad 225=64\left(3+2 x-x^{2}\right) \\
64 x^{2}-128 x+33=0
\end{gathered}
$$

The equation (5) can only be satisfied by the roots of this equation, namely $1+\sqrt{31} / 8$ and $1-\sqrt{31} / 8$. For the former, the left side of (5) is negative, since at $x=1$ it is 0, and we have seen that it decreases with increasing $x$. For the latter,

$$
\begin{gathered}
\sqrt{3-\left(1-\frac{\sqrt{31}}{8}\right)}=\frac{\sqrt{32+2 \sqrt{31}}}{4}=\frac{\sqrt{31+2 \sqrt{31}+1}}{4}=\frac{\sqrt{31}+1}{4}, \text { and } \\
\sqrt{\left(1-\frac{\sqrt{31}}{8}\right)+1}=\frac{\sqrt{31-2 \sqrt{31}+1}}{4}=\frac{\sqrt{31}-1}{4} .
\end{gathered}
$$

Thus, the equation (5) can only be satisfied by $x_{0}=1-\frac{\sqrt{31}}{8}$, and it is indeed satisfied by this value. The inequality in the problem, therefore, holds for those $x$ for which

$$
-1 \leqq x<1-\frac{\sqrt{31}}{8}
$$

and for no other values.

Horváth József (Esztergom, Temesvári Pelbárt g. III. o. t.)