I claim that $\boxed{M=2}$ By b) we have that $x_{n+1}\geq 3x_n-\sum\limits_{i=0}^{n-1} x_i$ If the equality does not hold, then $M$ will not be the greatest possible, therefore, we will assume the equality case for now on. We have the following: $$x_{n+2}=3x_{n+1}-\sum\limits_{i=0}^{n} x_i=3(3x_n-\sum\limits_{i=0}^{n-1} x_i)-x_n-\sum\limits_{i=0}^{n-1} x_i=8x_n-4\sum\limits_{i=0}^{n-1}x_i$$Thus, the sequence is increasing, therefore, for sufficiently large $k$, $\frac{x_{k+1}}{x_k}=1$ Finally, we have that $M=\lim_{x\rightarrow \infty} \frac{x_{n+1}}{x_n}=2$

Considering the equality, we obtain: $x_n=3x_{n+1}-x_{n+2}-3x_{n}+x_{n+1}\Rightarrow x_{n+2}-4x_{n+1}+4x_n=0\Rightarrow x_n=2^n(an+b).$ Solving it gives us $x_n=2^{n-1}(n+2)\Rightarrow \frac{x_{n+1}}{x_n}=2\cdot\frac{n+3}{n+2}$. $ \ $ But $\lim_{x\rightarrow \infty} 2\cdot\frac{n+3}{n+2}=2\implies \boxed{M\leq 2}$. Now we will prove by induction that $x_{n+1}> (x_{n}+\dots+x_0)$. The Base Case is trivial: $x_n\geq 3x_{n-1}-(x_{n-2}+\dots+x_0)>3x_{n-1}-x_{n-1}\Rightarrow \boxed{x_n>2x_{n-1}}$. $x_{n+1}\geq3x_n-(x_{n-1}+\dots+x_0)>2x_n>(x_{n}+\dots+x_0)$, completing the induction. But $x_{n+1}> (x_{n}+\dots+x_0)$ implies $x_{n+1}>2x_n\implies\boxed{M\geq 2}$. Thus $2\leq M\leq 2\implies \boxed{M=2} \ _{\blacksquare}$

This problem is from Greece. https://artofproblemsolving.com/community/c6h2953425p26447167