By the Arithmetic mean-Geometric mean inequality
$$3\sqrt[3]{a^4b^4c^4} \le a^4 + b^4 + c^4,$$
with equality iff $a = b = c$. Thus
$$3^{3/4}abc \le (a^4 + b^4 + c^4)^{3/4},$$
with equality iff $a = b = c$. Also, two applications of the Cauchy-Schwarz inequality tell us that
$$(a+b+c)^4 \le (3(a^2+b^2+c^2))^2 \le 27(a^4+b^4+c^4),$$
again with equality iff $a = b = c$, i.e.,
$$a+b+c \le 3^{3/4}(a^4+b^4+c^4)^{1/4},$$
with equality iff $a = b = c$. Combining these results we see that
$$3^{3/4}abc(a+b+c) \le (a^4+b^4+c^4)^{3/4}3^{3/4}(a^4+b^4+c^4)^{1/4} = 3^{3/4}(a^4+b^4+c^4).$$
with equality iff $a = b = c$. This is the desired result.