Thanks a lot for the solution. This is exactly what I am looking for. Could you please check the index $j$ in your solution? It does not appeared in the term inside the product. Maybe you can simplify your expression using Pochhammer symbol.

@Per Thanks a lot for your heads-up anyway. The sufficient conditions like Newton inequality on coefficients for real polynomials to have only real roots are much harder to find. It looks like that Sturm sequence method is the only candidate.

@Suvrit Thanks for the suggestion. I did a google search with key words as you suggested. Most of papers I found basically mentioned that binomial convolution will preserve properties like log-concavity/log-convexity. But this seems not directly related what I am looking for. Please refer to my previous comment. Thanks

Newton's inequality is $2ln(b_k)\geq \ln(b_{k-1})+ln(b_{k+1})$)

@Per I downloaded P. Branden's paper and R. Stanley's paper as you suggested. But I was not able to find the specific theorem that I was looking for: How much log-concavity of the positive real coefficients is needed in order for the corresponding real polynomials to have only zero roots? The sufficient conditions that I am looking for is in terms of inequalities like Newton's inequality (i.e., linear in $ln(b_k)$). Newton's inequality is $2lnb_k\geqlnb_{k-1}+lnb_{k+1}$)

@Brendan $b_1=1$ is fine. Actually $b_1$ can be absorbed into $x$. So the coefficients become $\binom{n}{i}b_i b_1^{-i}$ They satisfy the same recurrence relation. Thanks

@Brendan $b_0=b_1=1$ is OK. $b_0$ can be divided through, $b_1$ can be absorbed into x. So their values won't affect if the polynomials have real roots or not. This is true for all such polynomials. Thanks

@Per Thanks a lot for the comment. The coefficients $\binom{n}{i}b_i$ definitely satisfy Newton inequality so that the polynomials $p_n(x)$ satisfy the necessary condition to have real zeros. Could you please point me to a specific paper (or a specific theorem) by Petter Branden on arXiv? Thanks- Mike