Alternatively, if you simply want a reference, this is an application of the highly nontrivial Schmüdgen theorem (a.k.a. Schmüdgen's Positivstellensatz) 1. The theorem says that if you have a compact semialgebraic set $K \subset \mathbb{R}^n$ defined by inequalities $\{g_1 \geq 0, \dots, g_s \geq 0\}$, then any positive polynomial $f$ on $K$ belongs to the cone generated by $\{g_1, \dots, g_s\}$, i.e. sums with positive coefficients of products of the $g_i$'s (including powers) together with squares of arbitrary polynomials.

In your case, you have $g_1(x)=x$ and $g_2(x)=1-x$, and because you're in the one variable case, you should be able to get rid of the arbitrary squares (I might come back to this answer when I have more time).

1 K. Schmüdgen, The K-moment problem for compact semi-algebraic sets. Math. Ann. 289 (1991), no. 2, 203-206. (link)

Added Later: Note that Schmüdgen's result is only when $f|K>0$. If you relax to $f|K\geq 0$, it is not true any more. More surprisingly, Stengle points out in the MR that this result does not hold if you replace $\mathbb{R}$ by an arbitrary real closed field.

Later Still: See Vicki Power's answer for a reference more suitable for your specific question.

Alternatively, if you simply want a reference, this is an application of the highly nontrivial Schmüdgen theorem (a.k.a. Schmüdgen's Positivstellensatz) 1. The theorem says that if you have a compact semialgebraic set $K \subset \mathbb{R}^n$ defined by inequalities $\{g_1 \geq 0, \dots, g_s \geq 0\}$, then any positive polynomial $f$ on $K$ belongs to the cone generated by $\{g_1, \dots, g_s\}$, i.e. sums with positive coefficients of products of the $g_i$'s (including powers) together with squares of arbitrary polynomials.

In your case, you have $g_1(x)=x$ and $g_2(x)=1-x$, and because you're in the one variable case, you should be able to get rid of the arbitrary squares (I might come back to this answer when I have more time).

1 K. Schmüdgen, The K-moment problem for compact semi-algebraic sets. Math. Ann. 289 (1991), no. 2, 203-206. (link)

Added Later: Note that Schmüdgen's result is only when $f|K>0$. If you relax to $f|K\geq 0$, it is not true any more. More surprisingly, Stengle points out in the MR that this result does not hold if you replace $\mathbb{R}$ by an arbitrary real closed field.

Alternatively, if you simply want a reference, this is an application of the highly nontrivial Schmüdgen theorem (a.k.a. Schmüdgen's Positivstellensatz) [1]. The theorem says that if you have a compact semialgebraic set $K \subset \mathbb{R}^n$ defined by inequalities $\{g_1 \geq 0, \dots, g_s \geq 0\}$, then any positive polynomial $f$ on $K$ belongs to the cone generated by $\{g_1, \dots, g_s\}$, i.e. sums with positive coefficients of products of the $g_i$'s (including powers) together with squares of arbitrary polynomials.

In your case, you have $g_1(x)=x$ and $g_2(x)=1-x$, and because you're in the one variable case, you should be able to get rid of the arbitrary squares (I might come back to this answer when I have more time).

[1] K. Schmüdgen, The K-moment problem for compact semi-algebraic sets. Math. Ann. 289 (1991), no. 2, 203-206.

Alternatively, if you simply want a reference, this is an application of the highly nontrivial Schmüdgen theorem (a.k.a. Schmüdgen's Positivstellensatz) 1. The theorem says that if you have a compact semialgebraic set $K \subset \mathbb{R}^n$ defined by inequalities $\{g_1 \geq 0, \dots, g_s \geq 0\}$, then any positive polynomial $f$ on $K$ belongs to the cone generated by $\{g_1, \dots, g_s\}$, i.e. sums with positive coefficients of products of the $g_i$'s (including powers) together with squares of arbitrary polynomials.

In your case, you have $g_1(x)=x$ and $g_2(x)=1-x$, and because you're in the one variable case, you should be able to get rid of the arbitrary squares (I might come back to this answer when I have more time).

1 K. Schmüdgen, The K-moment problem for compact semi-algebraic sets. Math. Ann. 289 (1991), no. 2, 203-206. (link)

Added Later: Note that Schmüdgen's result is only when $f|K>0$. If you relax to $f|K\geq 0$, it is not true any more. More surprisingly, Stengle points out in the MR that this result does not hold if you replace $\mathbb{R}$ by an arbitrary real closed field.