Ankoganit wrote: A real polynomial of odd degree has all positive coefficients. Prove that there is a (possibly trivial) permutation of the coefficients such that the resulting polynomials has exactly one real zero. What ????

Sorry, that was a typo. Fixed now. Thanks.

alternating increasing (for odd powers) and decreasing (for even powers) works or the opposite, if I haven't messed up my memory...

Suppose the polynomial has degree $2n-1$ and the coefficients are, WLOG, $a_0<a_1<\cdots <a_{2n-1}$. We claim that the polynomial $$P(n)=\color{red}a_{2n-1}x^{2n-1}\color{black}+\color{blue}a_0x^{2n-2}\color{black}+\color{red}a_{2n-2}x^{2n-3}\color{black}+\color{blue}a_1x^{2n-3}\color{black}+\cdots +\color{red}a_{n-1}x\color{black}+\color{blue}a_n$$works just fine. Let's first state and prove a lemma. $\textbf{Lemma:}$ The function $f(x)=x^{2n-1}+x^{2n-2}+\cdots+x+1$ is monotonic increasing for any $n$. $\textbf{Proof:}$ After noticing that $$f(x)=\frac{x^{2n}-1}{x-1}\implies f'(x)=\frac{(2n-1)x^{2n}+1-2nx^{2n-1}}{(x-1)^2},$$it remains to show that $f'(x)\ge 0\iff (2n-1)x^{2n}+1\ge 2nx^{2n-1}$, which is vanilla AM-GM: $$\underbrace{x^{2n}+x^{2n}+\cdots +x^{2n}}_{2n-1\text{ times}}+1\ge |2nx^{2n-1}|\ge 2nx^{2n-1}.$$So we are good. $\square$ Now, time to get back to the main problem. By induction on $n$, we intend to prove that $P(n)$ defined above has exactly one root. Actually, we'll show something stronger: $P(n)$ (or any odd-degree polynomial with coefficients so arranged, for that matter) is monotonic increasing; then the oddness of degree would imply the existence of at least one root, and hence exactly one root. The base case $n=1$ is trivial enough; $P(n)$ becomes a boring linear function. Let's tackle the induction step. Note that: $$\begin{array}{rl}P(x)=& \quad a_0\left( x^{2n-1}+x^{2n}+\cdots +1\right)\\ &+\left(a_{2n-1}-a_0\right)\left(x^{2n-1}\right)\\ &+\bigg(\left(a_{2n-2}-a_0\right)x^{2n-3}+\left(a_{1}-a_0\right)x^{2n-4}+\cdots +\left(a_n-a_0\right)\bigg).\end{array}$$The first summand is increasing because of our lemma, the second is trivially so, and the third one has exactly the same arrangement of coefficients as $P(x)$, but has a smaller degree, so by inductive hypothesis, that one's increasing too. As the sum of all these three things, $P(x)$ is increasing as well, proving our claim. Hurray. $\blacksquare$