Problem 1. We say that a quadruple of nonnegative real numbers $(a, b, c, d)$ is balanced if

$$
a+b+c+d=a^{2}+b^{2}+c^{2}+d^{2}
$$

Find all positive real numbers $x$ such that

$$
(x-a)(x-b)(x-c)(x-d) \geqslant 0
$$

for every balanced quadruple $(a, b, c, d)$.

(Ivan Novak)
First Solution. We'll call any $x \in\langle 0, \infty\rangle$ satisfying the problem's condition great. Let $(a, b, c, d)$ be a balanced quadruple. Without loss of generality let $a \geqslant b \geqslant c \geqslant d$. We can rewrite the equation $a^{2}+b^{2}+c^{2}+d^{2}=a+b+c+d$ as

$$
\left(a-\frac{1}{2}\right)^{2}+\left(b-\frac{1}{2}\right)^{2}+\left(c-\frac{1}{2}\right)^{2}+\left(d-\frac{1}{2}\right)^{2}=1
$$

which implies $\left(a-\frac{1}{2}\right)^{2} \leqslant 1$, meaning that $a \leqslant \frac{3}{2}$.

6 points.

If we take $x \geqslant \frac{3}{2}$, the values of $x-a, x-b, x-c$ and $x-d$ are all nonnegative. Thus, any $x \geqslant \frac{3}{2}$ is great.

1 point.

If we take $(a, b, c, d)=\left(\frac{3}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$, then for any $x \in\left\langle\frac{1}{2}, \frac{3}{2}\right\rangle$ we have $(x-a)(x-b)(x-c)(x-d)<0$. Thus, no $x \in\left\langle\frac{1}{2}, \frac{3}{2}\right\rangle$ is great.

2 points.

If we take $(a, b, c, d)=(1,0,0,0)$, then for any $x \in\langle 0,1\rangle$ we have $(x-a)(x-b)(x-c)(x-d)<0$. Thus, no $x \in\langle 0,1\rangle$ is great.

1 point.

We conclude that a number $x$ is great if and only if $x \geqslant \frac{3}{2}$.