A chain of squares: a⁴ + b⁴ + c⁴ ≥ abc(a + b + c)

Idea. Prove a stronger intermediate inequality twice with $x^2+y^2+z^2\ge xy+yz+zx$.

Basic fact. For all real $x,y,z$,
$$x^2+y^2+z^2-xy-yz-zx=\frac12\left[(x-y)^2+(y-z)^2+(z-x)^2\right]\ge0.$$

Step 1. With $x=a^2$, $y=b^2$, $z=c^2$:
$$a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2.$$

Step 2. With $x=ab$, $y=bc$, $z=ca$: $xy=ab^2c$, $yz=bc^2a$, $zx=ca^2b$, so
$$a^2b^2+b^2c^2+c^2a^2\ge ab^2c+abc^2+a^2bc=abc(a+b+c).$$

Equality. Step 1 is an equality iff $a^2=b^2=c^2$, and step 2 iff $ab=bc=ca$. Together, $|a|=|b|=|c|$ and the signs agree, so $a=b=c$.

Lesson. Look for a chain of simpler true inequalities, each of which is "square $\ge0$" after a substitution such as $x=ab$.