Irska

By the Arithmetic mean-Geometric mean inequality $$3\sqrt[3]{a^4{b}^4{c}^4}\le a^4+b^4+c^4,$$ 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.