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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:

enter image description here

The result is as follows:

enter image description here

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]

enter image description here

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

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1 Answer 1

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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))

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