Actually not very easy. The key is if $P(x,y,z)$ is a symmetric polynomial, then $P(x_1,x_2,x_3)\in\mathbb Z.$ BTW, seems like teacher 周晓东 gave the bound $\frac 1{(2M+2)^2}$ on a lesson, so molom!

Let $P(x) = x^3+Cx+D$ be the monic polynomial with roots $x_1-x_2$, $x_2-x_3$, and $x_3-x_1$. Note that $$C = \sum_{\text{cyc}}(x_1-x_2)(x_2-x_3) = -\sum_{\text{cyc}}x_2^2+ \sum_{\text{cyc}}x_1x_2 = -a^2+3b,$$so $|C| \leqslant M^2+3M$. Also, $D$ is the square root of the cubic discriminant of $f$, an integer. Then if $D$ is nonzero, then $|D| \geqslant 1$. In this case, the derivative of $P$ is $3x^2 + C$, which over $-\frac12\leqslant x\leqslant \frac12$ has magnitude at most $|C|+1$. Since $P(0) = D$, its roots must be at least $\frac{|D|}{|C|+1} \geqslant \frac{1}{M^2+3M+1}$ away from the origin. If $D$ is zero, then the roots of $P$ are $0,\pm\sqrt{C}$, which clearly have magnitude $\gg\frac{1}{M^2+3M+1}.$ By definition $x_1-x_2$ is one of the roots of $P$, so it has magnitude $\geqslant \frac{1}{M^2+3M+1}$ as desired.

Really,nice.Thx