The second property should be: b) $ a_{1}^{2}+a_{2}^{2}+...+a_{n-1}^{2}\leq\frac{4\cdot min (a_{0}^{2},a_{n}^{2})}{n-1}$ (otherwise there's no meaning to the min)

I've corrected it.

this one already was posted here : http://www.mathlinks.ro/viewtopic.php?t=133243 question was: for which $ a_{i}$ next inequality holds: $ \sum_{i = 0}^{n}a_{i}x^{i} \geq 0$ for all $ x$

Please help me with this problem,I did not find any hints from the posts above.Thanks.

Sorry to revive this one.But pretty interesting problem.Hope my solution is correct. The second condition tells us to apply Cauchy-Schwartz.And so do we. $|a_{n-1}x^{n-1}+...+a_1x|\le \sqrt{a_{n-1}^2+...+a_1}\cdot \sqrt{x^{2n-2}+...+x^2}....(*)$.Now see that $P(x)=a_nx^n+a_0+(a_{n-1}x^{n-1}+...+a_1x)\ge \text{min}(a_0,a_n)(x^n+1)-|a_{n-1}x^{n-1}+...+a_1x|\ge \text{min}(a_0,a_n)(x^n+1)-\sqrt{a_{n-1}^2+...+a_1}\cdot \sqrt{x^{2n-2}+...+x^2}\ge \text{min}(a_0,a_n)\left(x^n+1-\frac{2\sqrt{x^{2n-2}+...+x^2}}{\sqrt{n-1}}\right)$. Here we used $(*)$ and the second condition.Now we will show that $(n-1)(x^n+1)^2\ge 4\sum_{i=1}^{n-1}x^{2i}$ we can put $n=2m$ and rewrite the equivalent inequality using $x^2=t$. $(2m-1)(t^{2m}+2t^m+1)\ge 4\sum_{i=1}^{2m-1}t^i$ which follows easily from Karamata .Since $f(u)=t^u$ is convex and the vectors ${a}=(2m-1,2m-1,2m-1,2m-1,2m-2,2m-2,2m-2,2m-2,....,1,1,1,1)$ where each of $2m-1,2m-2,....,1$ occurs $4$ times and ${b}=(2m,2m,....,2m,m,m,...,m,0,0,...,0)$ where each of $2m,m,0$ occur $2m-1,4m-2,2m-1$ times respectively; among ${a}$ and ${b}$ see that $b \succ a$ now Karamata gives us exactly the inequality we are asking for.