Assume otherwise. Wlog let $a \le b \le c$. Then $b < a+1, c <a+1$. Let $c= 1+a-e, b=1+a-d$, $1\ge d,e>0$. Then we have $a=\frac{d+e}{3}$, and thus the second condition becomes $d(d-1) + e(e-1) = de$, which is clearly false.

From the conditions we get $$2=3(a^2+b^2+c^2)-(a+b+c)^2=(a-b)^2+(b-c)^2+(c-a)^2.$$Let $a\geqslant b\geqslant c$. We have $$2(c-a)^2 \geqslant (a-b)^2+(b-c)^2+(c-a)^2 = 2 \Rightarrow |c-a| \geqslant 1.$$