Let $P(x)=a_{2n}x^{2n}+a_{2n-1}x^{2n-1}+\ldots+a_{0}$ be an even degree polynomial with positive coefficients.
Is it possible to permute the coefficients of $P(x)$ so that the resulting polynomial will have NO real roots.
Yes: put the $n+1$ largest coefficients on the even powers of $x$, and the $n$ smallest coefficients on the odd powers of $x$.
Clearly the polynomial will have no nonnegative roots regardless of the permutation. Changing $x$ to $-x$, it suffices to show: if $\min\{a_{2k}\} \ge \max\{a_{2k+1}\}$, then when $x>0$,$$a_{2n}x^{2n} - a_{2n-1}x^{2n-1} + \cdots + a_2x^2 -a_1x+a_0$$is always positive.
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