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5th power equation

Author: mathforces
Problem has been solved: 63 times

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Given 5 real numbers $a \leq b \leq c \leq d \leq e$ such that: $$\begin{cases} a+b+c+d+e=-1 \\ a^2+b^2+c^2+d^2+e^2=15 \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}=-1 \\ \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}+\frac{1}{e^2}=15 \\ abcde=-1 \end{cases} $$
Find second smallest integer value of $ ed$.

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