<s>[INST] Weird Maximization Problem For $1 \leq i \leq 215$ let $a_i = \dfrac{1}{2^{i}}$ and $a_{216} = \dfrac{1}{2^{215}}$. Let $x_1, x_2, \dots, x_{216}$ be positive real numbers such that $\displaystyle\sum_{i=1}^{216} x_i=1$ and
$$\sum_{1 \leq i < j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}.$$Find the maximum possible value of $x_2.$

I simplified the condition to $\displaystyle\sum_{i=1}^{216}\dfrac{x_i^2}{1-a_i}=\dfrac{1}{215},$ but I'm not sure what to do next.
 [/INST] Given $ \sum \frac{ x_i ^2 } { 1 - a_i} = \frac{1}{215}$, apply Cauchy Schwarz / Titu's lemma to get
$$ \sum \frac{ x_i ^2 } { 1 - a_i} \geq \frac{ ( \sum x_i) ^ 2 } { \sum 1 - a_i } = \frac{ 1^2 } { 215 } . $$
Since equality holds throughout, we conclude that $ \frac{ x_i } { 1-a_i}$ is a constant, say $k$.
Since $ 1 = \sum x_i = \sum (1-a_i)k  = 215 k$, so $ k = \frac{1}{215}$.
Hence, $ x_ i = \frac{ 1 - a_i } { 215 }$.
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