3. (9th "Hope Cup" Invitational Competition Question) If $0<a, b, c<1$, and $a+b+c=2$, then the minimum value of $a^{2}+b^{2}+c^{2}$ is

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3. Fill $\frac{4}{3}$. Reason: Note the identity $3\left(a^{2}+b^{2}+c^{2}\right)=(a+b+c)^{2}+(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$. Substituting the condition $a+b+c=2$, we get $3\left(a^{2}+b^{2}+c^{2}\right)=2^{2}+(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$, which means $3\left(a^{2}+b^{2}+c^{2}\right) \geqslant 4$. Therefore, the minimum value of $a^{2}+b^{2}+c^{2}$ is $\frac{4}{3}$.