Let's determine all numbers $p$ for which there exist $x, y$ such that $p=x+y$ and

$$
x^{2}+4 y^{2}+8 y+4 \leq 4 x
$$
$$
x^{2}+4 y^{2}+8 y+4 \leq 4 x
$$

Solution I. Suppose $p$ is a number for which there exist $x, y$ such that $p=x+y$ and the numbers $x, y$ satisfy the inequality (1). Then, substituting $(p-x)$ for $y$ in (1), we obtain the inequality

$$
5 x^{2}-4(2 p+3) x+4\left(p^{2}+2 p+1\right) \leq 0
$$

which has a solution for $x$. The discriminant of the quadratic polynomial on the left side of (2) must therefore be non-negative. This gives us the following condition for $p$:

$$
-p^{2}+2 p+4 \geq 0
$$

Since the coefficient of $p^{2}$ in the quadratic polynomial on the left side of (3) is negative, the value of $p$ must lie between the roots of the polynomial (including the roots):

$$
1-\sqrt{5} \leq p \leq 1+\sqrt{5}
$$

We will show that any $p$ satisfying the inequality (4) is a solution to this problem. If $p$ satisfies (4), then it also satisfies (3). This, in turn, implies that there exists an $x$ that satisfies the inequality (2). Such an $x$ and the number $y=p-x$ will also satisfy the inequality (1).

Therefore, the numbers $p$ that satisfy the inequality (4) and only these numbers satisfy the conditions of the problem.

Solution II. The inequality (1) can be transformed using the substitution $z=2 y$ to

$$
(x-2)^{2}+(z+2)^{2} \leq 4
$$

The points $(x, z)$ that satisfy inequality (5) lie on or inside the circle with center $(2, -2)$ and radius 2 in the plane.

![](https://cdn.mathpix.com/cropped/2024_05_02_597b70087999cd69acb8g-1.jpg?height=1323&width=631&top_left_y=1344&top_left_x=736)

The points $(x, z)$ that satisfy the equation $p=x+y$ (or equivalently $p=x+z/2$) lie on the line

$$
z=-2 x+2 p
$$

Based on this, the problem can be restated as finding all values of $p$ for which the line (6) has a common point with the circular region defined by inequality (5).

The lines (6) are parallel. We need to find the tangents to the circle among these lines - these still have a common point with the circular region - and the values of $p$ corresponding to these tangents will give the solution to the problem. The tangents to the circle

$$
(x-2)^{2}+(z+2)^{2} \leq 4
$$

with slope -2 can be determined by simple calculation, and the corresponding values of $p$ are: $p_{1}=1-\sqrt{5}$ and $p_{2}=1+\sqrt{5}$. Therefore, the solution to the problem is given by the values of $p$ that satisfy the inequality

$$
1-\sqrt{5} \leq p \leq 1+\sqrt{5}
$$

Balázs Iván József (Budapest, Fazekas M. Gyak. Gimn., IV. o. t.)