6. Find all triples of positive numbers $a, b, c$, satisfying the conditions $a+b+c=3, a^{2}-a \geqslant 1-b c$, $b^{2}-b \geqslant 1-a c, \quad c^{2}-c \geqslant 1-a b$.
Answer: $a=b=c=1$. If none of the numbers equals 1, then there are two on one side of 1, let these be $a$ and $b$. Adding the first two conditions and substituting $c=3-a-b$, we get $(a-1)(b-1) \leqslant 0$, a contradiction. If, however, $c=1$, then $a+b=2$ and from the third condition $ab \geqslant 1$, which only happens when $a=b=1$.