# Solving a system of inequalities involving Binomial

Is there a better method to solve this system of inequalities regarding the binomial distribution?

In mathematics, the description and result of this inequality are as shown in the picture below:

The result is as follows:

The following code cannot directly solve the solution set of the inequality.

Clear["*"]
eq1 = Binomial[n, k] (1 - p)^(n - k) p^k >=
Binomial[n, k + 1] (1 - p)^(n - k - 1) p^(k + 1);
eq2 = Binomial[n, k] (1 - p)^(n - k) p^k >=
Binomial[n, k - 1] (1 - p)^(n - k + 1) p^(k - 1);
Reduce[{eq1, eq2, 1 > 1 - p > 0, 1 > p > 0, n >= k >= 0,
n \[Element] Integers, k \[Element] Integers}, k]


It is necessary to transform each of the two inequalities separately before we can solve for the final solution set, as detailed below.

Clear["*"]
eq1 = Binomial[n, k] (1 - p)^(n - k) p^k >=
Binomial[n, k + 1] (1 - p)^(n - k - 1) p^(k + 1);
eq2 = Binomial[n, k] (1 - p)^(n - k) p^k >=
Binomial[n, k - 1] (1 - p)^(n - k + 1) p^(k - 1);
eq1a = MultiplySides[
eq1, (1 - p)^(-n + k + 1) p^(-k)/( Binomial[n, k]),
Assumptions -> 1 > p > 0 && n >= k >= 1] // FunctionExpand
eq2a = MultiplySides[
eq2, (1 - p)^(-n + k) p^(-k + 1)/(Binomial[n, k]),
Assumptions -> 1 > p > 0 && n >= k >= 1] // FunctionExpand
Reduce[{eq1a, eq2a, 1 > 1 - p > 0, 1 > p > 0, n >= k >= 0,
n \[Element] Integers, k \[Element] Integers}, k]


Is there a better method to solve this system of inequalities 14.1? I am using version 14.1.

The following works in both 14.0 and 14.1 on Windows 10.

Clear["*"]
eq1 = Binomial[n, k] (1 - p)^(n - k) p^k >=
Binomial[n, k + 1] (1 - p)^(n - k - 1) p^(k + 1);
eq2 = Binomial[n, k] (1 - p)^(n - k) p^k >=
Binomial[n, k - 1] (1 - p)^(n - k + 1) p^(k - 1);
Reduce[{FullSimplify[PowerExpand@FunctionExpand[eq1, eq2],
Assumptions -> {1 > 1 - p > 0, 1 > p > 0, n >= k >= 0}],
n >= k >= 0, 1 > 1 - p > 0, 1 > p > 0, n \[Element] Integers, k \[Element] Integers}, k]


Element[n | k, Integers] && 0 < p < 1 && ((n == 0 && k == 0) || (Inequality[0, Less, n, LessEqual, (1 - p)/p] && 0 <= k <= n) || (n > (1 - p)/p && -1 + p + n*p <= k <= n))`