Hölder's inequality for three sequences

Idea. Normalise, then use AM-GM on each triple.

Normalisation. Let $A=\sum a_i^3$, $X=\sum x_i^3$, $U=\sum u_i^3$. If one of them is $0$, both sides vanish. Otherwise, for each $i$ AM-GM on three non-negative numbers gives
$$\frac{a_i^3}A+\frac{x_i^3}X+\frac{u_i^3}U\ \ge\ 3\,\frac{a_ix_iu_i}{(AXU)^{1/3}}.$$

Summation. Summing over $i$, the left side is $3$, so
$$3\ge3\,\frac{\sum a_ix_iu_i}{(AXU)^{1/3}}\ \Longrightarrow\ \sum a_ix_iu_i\le(AXU)^{1/3},$$
which is the claim after cubing.

Consequence. Take $x_i=u_i=1$ and $n=3$: $(a^3+b^3+c^3)\cdot3\cdot3\ge(a+b+c)^3$, i.e. $a^3+b^3+c^3\ge\frac{(a+b+c)^3}9$.

Remark. Equality holds iff the three sequences $(a_i^3),(x_i^3),(u_i^3)$ are proportional.

Lesson. To prove a product-of-sums inequality, normalise every factor to $1$ and let AM-GM do the work term by term.