Serbian Mathematical Olympiad

Solution:

Let $\max \left\{a_{1}, a_{2}\right\} \leqslant a_{3} \leqslant a_{4}$. Denote $a_{2}=\beta^{2}$ and $a_{3}=\gamma^{2}, \beta, \gamma \geqslant 0$. From the condition of the problem it follows that $a_{1} \leqslant(\gamma-\beta)^{2}$ and $a_{4} \geqslant(\gamma+\beta)^{2}$.
Suppose that both of these inequalities are in fact equalities. Then we have $a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}=3\left(\beta^{4}+4 \beta^{2} \gamma^{2}+\gamma^{4}\right)$ and $a_{1} a_{2}+a_{1} a_{3}+a_{1} a_{4}+a_{2} a_{3}+a_{2} a_{4}+a_{3} a_{4}=3\left(\beta^{4}+\beta^{2} \gamma^{2}+\gamma^{4}\right)$. Moreover $\gamma \leqslant 2 \beta$, so
$$\frac{\beta^{4}+4 \beta^{2} \gamma^{2}+\gamma^{4}}{\beta^{4}+\beta^{2} \gamma^{2}+\gamma^{4}}=1+\frac{3 \beta^{2} \gamma^{2}}{\beta^{4}+\beta^{2} \gamma^{2}+\gamma^{4}}=1+\frac{3}{1+\frac{\beta^{2}}{\gamma^{2}}+\frac{\gamma^{2}}{\beta^{2}}} \geqslant \frac{11}{7}$$
with equality for $\gamma=2 \beta$. Therefore, in this case we have $K \geqslant \frac{11}{7}$, and equality is attained for $a_{1}: a_{2}: a_{3}: a_{4}=1: 1: 4: 9$.

Let us also show that we can take $a_{1}=(\gamma-\beta)^{2}$ and $a_{4}=(\gamma+\beta)^{2}$. Consider the expression
$$F=a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}-\frac{11}{7}\left(a_{1} a_{2}+a_{1} a_{3}+a_{1} a_{4}+a_{2} a_{3}+a_{2} a_{4}+a_{3} a_{4}\right)$$
For fixed $a_{2}, a_{3}, a_{4}, F$ is a decreasing function of $a_{1}$ for $a_{1}<\frac{11}{14}\left(a_{2}+a_{3}+a_{4}\right)$, where $\frac{11}{14}\left(a_{2}+a_{3}+a_{4}\right) \geqslant \frac{11}{14}\left(\beta^{2}+\gamma^{2}+(\beta+\gamma)^{2}\right) \geqslant(\gamma-\beta)^{2} \geqslant a_{1}$, so $F$ does not increase if we replace $a_{1}$ with $(\gamma-\beta)^{2}$. Now we may assume without loss of generality that $a_{1} \leqslant a_{2}$, i.e. $\beta \leqslant \gamma \leqslant 2 \beta$. Similarly as above, for fixed $a_{1}, a_{2}, a_{3}, F$ is an increasing function of $a_{4}$ for $a_{4}>\frac{11}{14}\left(a_{1}+a_{2}+a_{3}\right)$, and moreover $\frac{11}{14}\left(a_{1}+a_{2}+a_{3}\right) \leqslant \frac{11}{14}\left(\beta^{2}+\gamma^{2}+(\gamma-\beta)^{2}\right) \leqslant(\gamma+\beta)^{2} \leqslant a_{4}$, hence $F$ does not increase if we replace $a_{4}$ with $(\gamma+\beta)^{2}$.