This isn't too hard. If $ f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 $, then the given condition gives $ f(1)^2 = (\sum a_i )^2 \geq 1 $. So for any x, Cauchy gives \[ f(x) f(1/x) = (\sum a_i x^i) (\sum \frac{a_i}{x^i}) \geq (\sum a_i)^2 \geq 1. \] And since everything's positive, the result follows.

Yes, I know, this was one of the proposed solutions. Another one, not using Cauchy, goes as follows: With $P\left( x\right) =\sum_{k=0}^{n}a_{k}x^{n}$, where all $a_{i}$ are positive, we have $P\left( x\right) \cdot P\left( \frac{1}{x}\right) =\sum_{k=0}^{n}a_{k}x^{k}\cdot \sum_{k=0}^{n}\frac{a_{k}}{x^{k}}=\sum_{0\leq i\leq n,\;0\leq j\leq n}\left( a_{i}x^{i}\cdot \frac{a_{j}}{x^{j}}\right) $ $=\sum_{0\leq i\leq n,\;0\leq j\leq n}\left( a_{i}a_{j}x^{i-j}\right) =\sum_{i=0}^{n}a_{i}^{2}+\sum_{0\leq i<j\leq n}\left( a_{i}a_{j}x^{i-j}+a_{j}a_{i}x^{j-i}\right)$ $=\sum_{i=0}^{n}a_{i}^{2}+\sum_{0\leq i<j\leq n}\left( a_{i}a_{j}\left( x^{i-j}+\frac{1}{x^{i-j}}\right) \right) $ $\geq \sum_{i=0}^{n}a_{i}^{2}+\sum_{0\leq i<j\leq n}\left( 2a_{i}a_{j}\right) \text{\ \ \ \ \ \ \ \ \ \ (since }x^{i-j}+\frac{1}{x^{i-j}}\geq 2\text{)} $ $=\left( \sum_{k=0}^{n}a_{k}\right) ^{2}=\left( P\left( 1\right) \right)^{2}\geq 1$ (but I guess it's just the same). Darij

Sorry ...but who is Arthur Englel??

He is the Author of a book called "Problem-Solving Strategies", wich I think is very interesting, it helps a lot . Also I think he is (I'm not sure) one of the trainers of USA IMO team.

Darij wrote: Quote: (but I guess it's just the same) In fact you're right. What you exactly do is to show Cauchy in middle of the problem. I think it's just easier to assume it true and use it

Leonardo wrote: He is the Author of a book called "Problem-Solving Strategies", wich I think is very interesting, it helps a lot . Also I think he is (I'm not sure) one of the trainers of USA IMO team. Well, if you replace "USA" by "German", then it's completely right... The book is indeed very good, but unfortunately there are some errors ranging from typos up to incorrect proofs... but this is "comme il faut" nowadays. Darij

Can i find this book on the internet anywhere? Thx a lot... If it's interesting... it may surely help _______________________ "Our days are never coming back... " (Highway Song -> System of a down)

I think you'll have to buy it. It's quite expensive btw.

Ah.... that's a bad news for a Moldavian... It seems quite an interesting book at first sight... Palosh alias Downito ____________________________________ "Our days are never coming back... " (Highway Song -> System of a down)

darij grinberg wrote: Let P be a polynomial with positive coefficients. Prove: If the inequality $P\left( \frac{1}{x}\right) \geq \frac{1}{P\left( x\right) }$ is valid for x = 1, then it is valid for all x > 0, too. This is also Baltic Way 2004 Problem 2: Let P(x) be a polynomial with positive coefficients. Prove that if $P\left(\frac{1}{x}\right)P\left(x\right) \geq 1$ holds for x = 1, then the same inequality holds for each positive x. Darij

Palosh wrote: Can i find this book on the internet anywhere? Search for keyword '' Problem-Solving Strategies'' on Google , You will have that book!