Can we permute the coefficients of a polynomial so that it has NO real roots? 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.

* If $x\ge1$, this follows from $$ (a_{2n}x^{2n} - a_{2n-1}x^{2n-1}) + \cdots + (a_2x^2 -a_1x) +a_0 \ge 0 + \cdots + 0 + a_0 > 0\. $$
* If $0 0\. \end{multline*}