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Let $f$ be a real coefficient homogeneous polynomial in $n$ undeterminates, such that $f(x_1,\cdots,x_n)>0$ whenever $x_1,...,x_n$ are non-negative real numbers, not all $0$.

Then how to show that exists a natural number $N$ such that all coefficients of $(x_1+\cdots+x_n)^N f(x_1,\cdots,x_n)$ are strictly positive ?

I found that this problem was also asked in the AoPS forum, here, back in 2004, but without any proof or responses. I'm guessing this is a very old problem.

Any hints or ideas how to proceed ? Thank you.

I have another question, perhaps related .. is there a polynomial $f$ in $n$ undeterminates, such that $f(\bar{x}) > 0$, in $\mathbb{R}^n$ , but $\inf\limits_{\bar{x} \in \mathbb{R}^n} f = 0$.

r9m
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    The answer to the second question is yes; take $f(x, y) = x^2 + (xy - 1)^2$. See also the discussion at http://mathoverflow.net/questions/38019/zeros-of-gradient-of-positive-polynomials. – Qiaochu Yuan Jul 07 '14 at 02:54
  • @QiaochuYuan that's a nice example ... thanks :-) – r9m Jul 07 '14 at 02:55
  • Can this be solved for n=1 or 2? – marty cohen Jul 07 '14 at 03:58
  • @marty are you suggesting induction on $m$ ? – r9m Jul 07 '14 at 04:24
  • No. Perhaps seeing what happens in these simplest cases might suggest what to do for the general case. – marty cohen Jul 07 '14 at 14:19
  • Suppose all the $x$'s except one are zero. Then we know $f(0,\dots,x_i,\dots,0) >0$ whenever $x_i >0$. Thus the coefficient of $(x_i)^n$ in f must be positive. – DavidButlerUofA Jul 11 '14 at 00:39
  • I am not sure about the question. But consider a simple case:$$ f(x,y) = 2x+y$$ then $$(x+y)^N(2x+y)$$ can still be negative and the coefficients are alread positive? Can you explain more about your question? – johannesvalks Jul 12 '14 at 02:59
  • But I think that there is no such $N$.

    Consider $$f(x,y) = x^2 + y^2 - 2xy$$

    You can try $$(x+y)^N f(x,y)$$ but there always remain negative coefficients...

     – johannesvalks Jul 12 '14 at 03:21
  • Besides, the function $f(x_1,\cdots,x_m) = x_1^n + \cdots + x_n^n - n x_1 x_2 $ does not satisfy the condition $f(x_1,\cdots,x_n)>0$, for $x_k>0$, as $f(x,\cdots,x) = 0$. – johannesvalks Jul 12 '14 at 03:26
  • @johannesvalks sorry (I'm deleting the above comment) .. $f(x_1,x_2,\cdots,x_n) = x_1^n + x_2^n + \cdots +x_n^n - n x_1 x_2\cdots x_n$, vanishes if $x_1=x_2= \cdots =x_n$, in ${(x_1,x_2,\cdots,x_n)\in \mathbb{R}^n | x_i \ge 0, x_1+x_2+\cdots+x_n >0 }$, a proper example would be $x^4-x^2y^2 +y^4$ – r9m Jul 12 '14 at 03:27
  • @David: In general we also have $f(x,x,\cdots,x)>0$, so the $\sum a_\bar{k} > 0$. – johannesvalks Jul 12 '14 at 03:34
  • @r9m: Ok, thank you for the input, I now understand the question... – johannesvalks Jul 12 '14 at 17:32
  • Just adding things that may or may not help: if $f(x_1,...,x_m)$ factorises into (linear polynomial)*(something else), then the linear polynomial is not allowed to have negative coefficients, as otherwise it is possible for f to be zero when not all the $x_i$ are zero. – DavidButlerUofA Jul 14 '14 at 22:43
  • What would happen if you fixed $x_2$, ..., $x_n$ at particular non-negative values, not all zero? Then f would become an ordinary polynomial in $x_1$, which would have positive leading coefficient and it would also have to be strictly positive for non-negative $x_1$. This would imply that any zeros of this new polynomial would have to be strictly negative. Indeed, you could factorise it to $f(x_1) = a(x+\alpha_1)\dots(x+\alpha_r)g(x)$, where $g(x)$ is a product of monic irreducible quadratics and $\alpha_1$, ..., $\alpha_r$ are positive numbers. – DavidButlerUofA Jul 14 '14 at 23:06

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