How prove this $p(x)>0$ if $p(x)=\sum_{i=0}^{n}\binom{n}{i}a_{i}x^i(1-x)^{n-i}$ 
let the polynomials $$p(x)=\sum_{i=0}^{n}\binom{n}{i}a_{i}x^i(1-x)^{n-i}$$
and such
  $$a_{0}+\sum_{a_{i}<0}(1-\dfrac{i}{n})\binom{n}{i}a_{i}>0$$
  and
  $$a_{n}+\sum_{a_{i}<0}\dfrac{i}{n}\binom{n}{i}a_{i}>0$$
show that

$$p(x)>0,\forall x\in [0,1]$$
This problem is from a china analysis problem book excise, when this book introduce
 Bernstein Polynomials  give this hard problem,and I post

 A: From the conditions we know that $a_0>0$ and $a_n>0$, so the sum $\sum\limits_{a_i<0}$ excludes $a_0$ and $a_n$.
For $1\le i\le n-1$, by AM-GM,
$$
x^i(1-x)^{n-i} = \big(x^n\big)^{\frac in} \big((1-x)^n\big)^{\frac{n-i}n}
\le \frac in x^n + \frac{n-i}n (1-x)^n
= \frac in x^n + \left(1-\frac in\right) (1-x)^n.
$$
Then
$$
p(x) 
\ge a_0(1-x)^n + a_nx^n + \sum_{a_i<0} a_i\binom ni x^i(1-x)^{n-i}
\\\ge a_0(1-x)^n + a_nx^n + \sum_{a_i<0} a_i\binom ni \left(\frac in x^n + \left(1-\frac in\right) (1-x)^n\right)
= (1-x)^n\left(a_0+\sum_{a_i<0} \left(1-\frac in\right)\binom ni a_i\right)
+ x^n \left(a_n+\sum_{a_i<0} \frac in\binom ni a_i\right)
>0.
$$
A: To start, we have from the conditions that 
$$
p(0) = a_0 > a_0 + (\mathrm{negative~stuff}) > 0
$$
and
$$
p(1) = a_n > a_n + (\mathrm{negative~stuff}) > 0
$$
So at the endpoints of $p$, we are above zero.  What this is really saying is that $p$ doesn't have enough negative coefficients to overcome the two endpoints when $x = 0,1$.
The rest is essentially a convexity question.  I haven't worked out the details, but the two assumptions of this question relate to the local maxima of the individual Bernstein polynomials that make up $p(x)$, and that these maxima are not enough to overpower the two Bernstein polynomials at the endpoints, $B_{0,n}$ and $B_{n,n}$, and so the overall polynomial cannot become negative.
http://mathworld.wolfram.com/BernsteinPolynomial.html
