How prove this $p(x)>0$ if $p(x)=\sum_{i=0}^{n}\binom{n}{i}a_{i}x^i(1-x)^{n-i}$

Question

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

Answer 1

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 $$ p(x) \ge a_0(1-x)^n + a_nx^n + \sum_{a_i<0} a_i\binom ni x^i(1-x)^{n-i} \\\ge a_0(1-x)^n + a_nx^n + \sum_{a_i<0} a_i\binom ni \left(\frac in x^n + \left(1-\frac in\right) (1-x)^n\right) = (1-x)^n\left(a_0+\sum_{a_i<0} \left(1-\frac in\right)\binom ni a_i\right) + x^n \left(a_n+\sum_{a_i<0} \frac in\binom ni a_i\right) >0. $$

Answer 2

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