$\sum_{i=1}^8x_i=4,\sum_{1\leq i<j\leq 8}x_ix_j=7$. $\sum_{i=1}^8x_i^2=(\sum_{i=1}^8x_i)^2-2\sum_{1\leq i<j\leq 8}x_ix_j=2$ And $\sqrt{\frac{1}{8}\sum_{i=1}^8x_i^2}\geq\frac{1}{8}\sum_{i=1}^8x_i$ So $\sum_{i=1}^8x_i^2\geq2$,so the equality must holds and all $x_i$ is the same So $x_1=x_2=\cdots=x_8=\frac{1}{2}$,$f=\prod_{i=1}^8x_i=(\frac{1}{2})^8=\frac{1}{256}$

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Here is my solution: Let $S_n {=}^{def} \sum_{i=1}^n {r_i}^n$, where $r_i$ is a root of $P(x)$. Then By Newton's Identities, we have that: $\longrightarrow S_1 - 4=0 \implies S_1=4$ $\longrightarrow S_2 - 4S_1 + 14 =0 \implies S_2=2.$ Hence $S_2 = \sum_{i=1}^8 {r_i}^2$ From $ QM \ge AM$, we have : $\sqrt{\dfrac{1}{8} \sum_{i=1}^8 {r_i}^2} \ge \dfrac{1}{8} \sum_{i=1}^8 {r_i}$ $\implies \dfrac{1}{2} \ge \dfrac{1}{2} \iff r_i = r_j \forall i,j \in {1,2,\dots,8}$ Hence $f = \prod_{i=1}^8 r_i = \dfrac{1}{2^8} = \dfrac {1}{256}$

First note by the formula of $(x_1+x_2 + \cdots + x_8)^2$, that ${x_1}^2 + {x_2}^2 + \cdots + {x_8}^2 = 2$. By RMS $\geq$ AM inequality on $x_1, x_2, \cdots x_8$ gives that $RMS = AM$ which implies all the $x_i$'s are equal, to, say $r$. Hence $8r = 4 \implies r = \frac{1}{2}$ and $f = (\frac{1}{2})^8$

first of all we have $\sum_{i=1}^{8} x_{i}=4$ and $\sum_{i \neq j}x_{i} \cdot x_{j}=7$ so we have $\sum_{i=1}^{8} x_{i}^2=2$ also from cauchy swarchz we have we have $\sum_{i=1}^{8} x_{i}^2 \geqslant 2$ which gives equality case and hence $f=\frac{1}{256}$

Vieta gives $\sum x_i=4$ and $\sum x_i^2=2$, and since $x^2$ is convex, $\sum x_i^2\geq 2$, so equality must hold so $x_i=1/2$ for all $i$ so the answer is $1/256$.