China Mathematical Competition

Let $x=\frac{2}{3}a$. Then we have $|a|\le \frac{1}{3}$. Therefore (B) and (D) are excluded. By symmetry, (C) is also excluded. Then only (A) can be correct.

In general, for any $k\in \mathbb{R}$, let $x=\frac{1}{2}ka$. Then the original inequality becomes $$|a|\cdot |k-1|+\frac{3}{2}|a|\cdot \left|k-\frac{4}{3}\right|\ge |a{|}^2.$$ This is equivalent to $$|a|\le |k-1|+\frac{3}{2}\left|k-\frac{4}{3}\right|.$$ We have $$|k-1|+\frac{3}{2}\left|k-\frac{4}{3}\right|=\{\begin{array}{ll}\frac{5}{2}k-3,& k\ge \frac{4}{3},\\ 1-\frac{1}{2}k,& 1\le k<\frac{4}{3},\\ 3-\frac{5}{2}k,& k<1.\end{array}$$ So $$\underset{k\in \mathbb{R}}{\min }\left\{|k-1|+\frac{3}{2}\left|k-\frac{4}{3}\right|\right\}=\frac{1}{3}.$$ The inequality is reduced to $|a|\le \frac{1}{3}$. Answer: A.