It seems straightforward. Choose $x_i=\frac{a_i}{a_1+a_2+\ldots+a_n}$ then it becomes C.S.

It also looks like it's the only such sequence !?! EDIT. Indeed, this is the case! Let us denote $a = \sum_{i=1}^n a_i$, and take in particular $y_i = \dfrac {a_i} {a}$ for all $1\leq i\leq n$. Then \[a - \sum_{i=1}^n \frac{a_ix_i}{x_i+y_i} = a\sum_{i=1}^n x_i - \sum_{i=1}^n \frac{aa_ix_i}{ax_i+a_i} = a\sum_{i=1}^n\left (x_i - \frac{a_ix_i}{ax_i+a_i} \right) = a^2\sum_{i=1}^n \frac{x_i^2}{ax_i+a_i}\geq a^2 \frac{\left (\sum\limits_{i=1}^n x_i \right )^2}{a\sum\limits_{i=1}^n x_i+\sum\limits_{i=1}^n a_i} = a^2\dfrac {1} {a+a} = \dfrac {1} {2} a.\] This writes as \[\sum_{i=1}^n \frac{a_ix_i}{x_i+y_i} \leq \dfrac {1} {2} a,\] which is precisely our required inequality, reversed. Thus, for the required inequality to hold, we need equality in the above, which occurs if and only if $\frac{x_i}{ax_i+a_i} = \frac{x_j}{ax_j+a_j}$, i.e. $\frac{x_i}{a_i} = \frac{x_j}{a_j}$ for all $1\leq i,j\leq n$ (by the known C-S equality case). Let us denote by $x$ this common value; then $x_i = xa_i$ for all $1\leq i\leq n$, therefore $1 = \sum_{i=1}^n x_i = x\sum_{i=1}^n a_i = xa$. This means $x=\dfrac {1} {a}$, and so $x_i = \dfrac {a_i} {a}$ for all $1\leq i\leq n$, which thus is the unique $x$-sequence satisfying the required inequality for all eligible $y$-sequences.

we claim that $x_i=\frac{a_i}{a_1+....+a_n}$ for all $i=1,2,...,n$ works. Proof -- obvious by Titu's lemma

I used to try Cauchy as it's able to kill $y_i$ easily Click to reveal hidden text

but it's just too simply overshooted... And I always thought the answer is $\frac{1}{n}$ which is simply wrong...

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I died just looking at this.