is much cleaner, and requires only the two term rearrangement inequality $(x_iy_i+x_jy_j\le x_iy_j+x_jy_i)$:
For n=1 we have equality.
Let $n\ge2$ and assume that $(x_1+\cdots+x_n)(y_1+\cdots+y_n)\le n(x_1y_1+\cdota+x_ny_n)$;
then
$[(x_1+\cdots+x_n)+x_{n+1}][(y_1+\cdots+y_n)+y_{n+1}]
=(x_1+\cdots+x_n)(y_1+\cdots+y_n)
+\sum_{k=1}^n(x_ky_{n+1}+x_{n+1}y_k)
+x_{n+1}y_{n+1}$;
on the right hand side
the first term is $\le n(x_1y_1+\cdots+x_ny_n)$
by assumption,
since $x_ky_{n+1}+x_{n+1}y_k\le x_ky_k+x_{n+1}y_{n+1}$,
the second term is $\le (x_1y_1+\cdots+x_ny_n)+nx_{n+1}y_{n+1}$;
together with the third term, this makes the right side
$\le (n+1)(x_1y_1+\cdots+x_{n+1}y_{n+1}), and the proof is complete.
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