Berkeley Math Circle: Monthly Contest 1

Solution:
Since $a + b + c = 1$, we can multiply the right side by $a + b + c$ to get the equivalent inequality
$$a^{4} + b^{4} + c^{4} \geq a^{2} b c + a b^{2} c + a b c^{2} .$$
By AM-GM,
$$\frac{2 a^{4} + b^{4} + c^{4}}{4} = \frac{a^{4} + a^{4} + b^{4} + c^{4}}{4} \geq \sqrt[4]{a^{4} a^{4} b^{4} c^{4}} = a^{2} b c .$$
Similarly,
$$\begin{aligned} & \frac{a^{4} + 2 b^{4} + c^{4}}{4} \geq a b^{2} c \\ & \frac{a^{4} + b^{4} + 2 c^{4}}{4} \geq a b c^{2} \end{aligned}$$
Adding these three inequalities proves the desired inequality.