China Western Mathematical Olympiad

If $a=b=c=d=-1$, then $-3\ge k\cdot (-4)$. Hence $k\ge \frac{3}{4}$.

If $a=b=c=d=\frac{1}{2}$, then $4\cdot \frac{1}{8}+1\ge k\cdot (4\cdot \frac{1}{2})$. Thus $k\le \frac{3}{4}$, and so $k=\frac{3}{4}$.

At first, we prove that $4x^3+1\ge 3x$, $x\in [-1,+\infty )$. In fact, from $(x+1)(2x-1{)}^2\ge 0$, we have $4x^3+1\ge 3x$, $x\in [-1,+\infty )$. Therefore $$4a^3+1\ge 3a,$$ $$4b^3+1\ge 3b,$$ $$4c^3+1\ge 3c,$$ $$4d^3+1\ge 3d.$$ By adding the above 4 inequalities together, we get the inequality (1).