4. Find the maximum value of the expression $a+b+c+a b c$, where $a, b$ and $c$ are non-negative numbers satisfying $a^{2}+b^{2}+c^{2}+a b c \leqslant 4$.
4. The condition of the problem cannot hold if $a^{2}+b^{2}+c^{2}<3$: indeed, then by the A-G inequality $3(a b c)^{2 / 3} \leqslant a^{2}+b^{2}+c^{2}<3$, i.e., $a b c<1$, so $a^{2}+b^{2}+c^{2}+a b c<4$. Therefore, $a^{2}+b^{2}+c^{2} \geqslant 3$, but then $a+b+c \leqslant \sqrt{3\left(a^{2}+b^{2}+c^{2}\right)} \leqslant$ $a^{2}+b^{2}+c^{2}$, so $a+b+c+a b c \leqslant 4$. The last inequality is an equality for $a=b=c=1$, so the maximum possible value is 4.

Note. Under the condition $a^{2}+b^{2}+c^{2}+a b c=4$, the minimum possible value of the expression $a+b+c+a b c$ is 2.