66th Belarusian Mathematical Olympiad

$c_0,c_1,\dots ,c_{2n+1}$ of the numbers $a_0,a_1,\dots ,a_{2n+1}$ such that $c_{2n+1}=b_{2n+1}$, $c_{2n-1}=b_{2n-1}$, $\dots$, $c_1=b_{n+1}$ and $c_2=b_n$, $c_{2n-2}=b_{n-1}$, $\dots$, $c_0=b_0$. For the set of $c_0,c_1,\dots ,c_{2n+1}$ we have $$c_0\le c_1\ge c_2\le c_3\ge \cdots \ge c_{2n}\le c_{2n+1}.$$ Consider the polynomial $q(x)=c_{2n+1}{x}^{2n+1}+c_{2n}{x}^{2n}+c_{2n-1}{x}^{2n-1}+\cdots +c_{1}x+c_0$. Since the degree of $q(x)$ is odd and its leading coefficient is positive, we have $q(-x)<0$ and $q(x)>0$ for $x$ large enough. Therefore, $q(x)$ has at least one real root. If we show that $q(x)$ strictly increases, then $q(x)$ has at most one real root as required. It suffices to prove that the derivative of this polynomial is positive for all $x$. We have $$q^{\prime }(x)=(2n+1)c_{2n+1}{x}^{2n}+2n{c}_{2n}{x}^{2n-1}+\cdots +2c_{2}x+c_1.$$ Since $$\begin{array}{rl}& k{c}_{2k+1}{x}^{2k}+2k{c}_{2k}{x}^{2k-1}+k{c}_{2k-1}{x}^{2k-2}\ge \\ & \ge k{c}_{2k}(|x{|}^{2k}-2|x{|}^{2k-1}+|x{|}^{2k-2})=k{c}_{2k}|x{|}^{2k-2}(|x|-1{)}^2\ge 0\end{array}$$ for all $k\in \{1,2,\dots ,n\}$ and real $x$, we obtain $$\begin{array}{rl}{q}^{\prime }(x)& =(n+1)c_{2n+1}{x}^{2n}+(n{c}_{2n+1}{x}^{2n}+2n{c}_{2n}{x}^{2n-1}+n{c}_{2n-1}{x}^{2n-2})+\\ & ((n-1)c_{2n-1}{x}^{2n-2}+2(n-1)c_{2n-2}{x}^{2n-3}+(n-1)c_{2n-3}{x}^{2n-4})+\cdots +\\ & +(c_3{x}^2+2c_{2}x+c_1)\ge \\ & \ge (n+1)c_{2n+1}{x}^{2n}>0\end{array}$$ for $x\ne 0$. Since $q^{\prime }(0)=c_1>0$, we obtain $q^{\prime }(x)>0$ for all real $x$.