$\max _{p_i} \sum_{i=1}^n p_i^{(1-p_i)^a} $ s.t. $p_i \in [0,1]$ and $\sum_{i=1}^n p_i = 1$

Question

Consider the following problem $$ \max _{p_i} \sum_{i=1}^n p_i^{(1-p_i)^a} $$ s.t. $p_i \in [0,1]$ and $\sum_{i=1}^n p_i= 1$ for some given $a>0$. In other words, ${\bf p} = \{ p_i\}$ is a probability vector.

Question: Can we find ${\bf p}$ that maximizes the problem? If not, could we find a 'good' upper bound?

For the case of $n=2$, we have to maximize \begin{align} p^{(1-p)^a} + (1-p)^{p^a}, \end{align} which, from simulations, occurs at $p=1/2$ for all $a>0$.

Edit: Originally, I thought that the maximum occurs by choosing $p_i=1/n$. However, this appears not to be the case as given by the example in the chat for $n=3$.

Answer 1

Some thoughts.

Here we deal with the case $0 < a \le 2$. Let $f(x) := x^{(1-x)^a}$.

Fact 1. $f(x)$ is convex on $[0, c]$, and concave on $[c, 1]$, for some $c \in (0, 1)$.
(The proof is given at the end.)

Now we proceed. WLOG, assume that $p_1 \le p_2 \le \cdots \le p_n$. Let $$F(p_1, p_2, \cdots, p_n) := f(p_1) + f(p_2) + \cdots + f(p_n).$$ Then $F$ is maximal for $p_1 = \cdots = p_{k-1} = 0$ and $p_{k+1} = p_{k+2} = \cdots = p_n$ for some $k \in \{1, 2, \cdots, n\}$. The proof is easy. But many year ago, I knew Bin Zhao's half concave half convex theorem. So see the proof there.

(Edit 2024/12/03) Theorem 1. Let $a, b, c$ be fixed reals ($a < c < b$), and let $x_1, x_2, \cdots, x_n \in [a, b]$ ($n\ge 2$) with $x_1 + x_2 + \cdots + x_n = S$ (constant). Let $f$ be a continuous function on $[a, b]$, convex on $[a, c]$ and concave on $[c, b]$. Then $f(x_1) + f(x_2) + \cdots + f(x_n)$ is maximal for $x_1 = \cdots = x_{k-1} = a$ and $x_{k+1} = \cdots = x_n$ for some $k \in \{1, 2, \cdots, n\}$.

Proof of Theorem 1.

Let $Q$ be the set of all maximizers. Let $M(x_1, \cdots, x_n) := \#\{i : a < x_i < c, 1 \le i \le n\}$. Let $$p := \min_{(x_1, \cdots, x_n) \in Q} M(x_1, \cdots, x_n).$$

We claim that $p \le 1$. Assume, for the sake of contradiction, that $p \ge 2$. Let $(x_1, \cdots, x_n) \in Q$ with $M(x_1, \cdots, x_n) = p$. Then there exist $i < j$ such that $x_i, x_j \in (a, c)$.

(1) If $x_i + x_j - c \ge a$, by Karamata's inequality, we have $$f(x_i) + f(x_j) \le f(x_i + x_j - c) + f(c).$$ If we replace $x_i, x_j$ with $x_i + x_j - c, c$, we have a strictly smaller value of $M$.

(2) If $x_i + x_j - c < a$, by Karamata's inequality, we have $$f(x_i) + f(x_j) \le f(x_i + x_j - a) + f(a).$$ If we replace $x_i, x_j$ with $x_i + x_j - a, a$, we have a strictly smaller value of $M$.

For both cases, we get a contradiction. The claim is proved.

Now, WLOG, assume that $a \le x_1 \le x_2 \le \cdots \le x_n \le b$. Then there exists $k \in \{1, 2, \cdots, n\}$ such that $x_1 = \cdots = x_{k-1} = a$ and $c \le x_{k+1} \le \cdots \le x_n$. Also, we have $$f(x_{k+1}) + f(x_{k+2}) + \cdots + f(x_n) \le (n-k)f\left(\frac{x_{k+1} + \cdots + x_n}{n-k}\right)$$ with equality if $x_{k+1} = \cdots = x_n$. The desired result follows.

$\phantom{2}$

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Cleary $k < n$ (when $k=n$, $(p_1, \cdots, p_n) = (0, \cdots, 0, 1)$ which is not a maximizer). Let $p_k = y$. Then $y_{k+1} = \cdots = y_n = \frac{1-y}{n-k}$. We have $$F = y^{(1-y)^a} + (n-k)\left(\frac{1-y}{n-k}\right)^{\left(1-\frac{1-y}{n-k}\right)^a}.$$

Can we proceed?

$\phantom{2}$

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Proof of Fact 1.

We have $$f''(x) = x^{(1-x)^a}(1-x)^a g(x)$$ where \begin{align*} g(x) &:= \left(- \frac{a\ln x}{1-x} + \frac{1}{x}\right)^2(1-x)^a + \frac{a^2\ln x}{(1-x)^2} - \frac{a\ln x}{(1-x)^2} - \frac{2a}{x(1-x)} - \frac{1}{x^2}. \end{align*}

We can prove that $g'(x) < 0$ on $(0, 1)$ (the proof is given at the end).

Note that $\lim_{x\to 0^{+}} g(x) = \infty$ and $\lim_{x\to 1^{-}} g(x) = -\infty$. Thus, there exists $c \in (0, 1)$ such that $g(x) > 0$ on $(0, c)$ and $g(x) < 0 $ on $(c, 1)$. Thus, there exists $c \in (0, 1)$ such that $f''(x) > 0$ on $(0, c)$ and $f''(x) < 0 $ on $(c, 1)$. The desired result follows.

$\phantom{2}$

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Proof of $g'(x) < 0$.

After some manipulations, it suffices to prove that, for all $0 < x < 1$, $$M (1-x)^a + N < 0,$$ where ($Q := \ln x$) \begin{align*} M &:= -(1 - x - Qax)[a(2-a)x^2Q + (1-x)(3ax-2x+2)],\\ N &:= 2\,{a}^{2}Q{x}^{3}-2\,aQ{x}^{3}-{a}^{2}{x}^{3}+{a}^{2}{x}^{2}+5\,a{x}^ {3}-7\,a{x}^{2}-2\,{x}^{3}\\ &\qquad +2\,ax+6\,{x}^{2}-6\,x+2. \end{align*} It is easy to prove that $M \le 0$.

If $0 < a \le 1$, by Bernoulli inequality, we have $(1-x)^a = \frac{1-x}{(1-x)^{1-a}} \ge \frac{1-x}{1 - x(1-a)}$, and it suffices to prove that $$M \cdot \frac{1-x}{1 - x(1-a)} + N < 0, $$ or $$m_2 a^2 + m_1 a + m_0 < 0,$$ where \begin{align*} m_2 &= {Q}^{2}{x}^{3}-{Q}^{2}{x}^{2}+2\,Q{x}^{3}-{x}^{3}+{x}^{2},\\ m_1 &= -2\,{Q}^{2}{x}^{3}+2\,{Q}^{2}{x}^{2}-6\,Q{x}^{2}+6\,{x}^{3}+4\,Qx-9\,{ x}^{2}+3\,x,\\ m_0 &= -2\,Q{x}^{3}+8\,Q{x}^{2}-4\,{x}^{3}-8\,Qx+9\,{x}^{2}+2\,Q-6\,x+1. \end{align*} We can prove that $m_2, m_1, m_0 < 0$ on $(0, 1)$. The desired result follows.

If $1 < a \le 2$, by Bernoulli inequality, we have $(1-x)^a = \frac{(1-x)^2}{(1-x)^{2-a}} \ge \frac{(1-x)^2}{1-x(2-a)}$, and it suffices to prove that $$M \cdot \frac{(1-x)^2}{1-x(2-a)} + N < 0,$$ or $$q_3a^3 + q_2a^2 + q_1 a + q_0 < 0, \tag{B1}$$ where \begin{align*} q_3 &= -{Q}^{2}{x}^{4}+2\,{Q}^{2}{x}^{3}-{Q}^{2}{x}^{2}+2\,Q{x}^{3}-{x}^{3}+{ x}^{2},\\ q_2 &= 2\,{Q}^{2}{x}^{4}-4\,{Q}^{2}{x}^{3}-4\,Q{x}^{4}+2\,{Q}^{2}{x}^{2}+6\,Q {x}^{3}-10\,Q{x}^{2}\\ &\qquad +7\,{x}^{3}+4\,Qx-10\,{x}^{2}+3\,x,\\ q_1 &= 4\,Q{x}^{4}-10\,Q{x}^{3}-3\,{x}^{4}+16\,Q{x}^{2}-10\,Qx+7\,{x}^{2}+2\, Q-5\,x+1,\\ q_0 &= 2\,{x}^{4}-6\,{x}^{3}+6\,{x}^{2}-2\,x. \end{align*} (B1) is true (the proof is smooth).

We are done.

Answer 2

Note: It appears my previous answer wasn't very rigorous and had gaps in its reasoning (especially with $n-1$ EV). I've edited this post with some more progress I've made on this problem.

Going off River Li's answer, we know the maximum for $0< a\leq 2$ occurs when $p_k=y$, $p_1=p_2=\dots=p_{k-1}=\frac{1-y}{k-1}\geq y$, and $p_{k+1}=p_{k+2}=\dots=p_n=0$. (I've slightly changed the indexing method for convenience). Now, this simplifies the problem into maximizing $$g(y)=y^{(1-y)^a}+(k-1)\left(\frac{1-y}{k-1}\right)^{\left(1-\frac{1-y}{k-1}\right)^a},$$ for $y\leq\frac 1k$. Then, we have the following claim:

Claim: If $g''(y)$ is convex on $(0,c)$ and concave on $(c,1)$ for some $c$, then the maximizing solution is in the form $p_1=\dots=p_k=\frac 1k$ and $p_{k+1}=\dots =p_n=0$.

Remark: The convex-concave property of $g''(y)$ appears to hold numerically, but I'm still working on a proof for this. Any ideas are appreciated.

Proof of Claim. We can write $g(y)=f(y)+(k-1)f\left(\frac{1-y}{k-1}\right)$ for $f(y)=y^{(1-y)^a}$. Now, note that $g'(y)=f'(y)-f'\left(\frac{1-y}{k-1}\right)$, which has a zero at $y=\frac{1}{k}$.

Now, assume $g''(y)=f''(y)+\frac{1}{k-1}f''\left(\frac{1-y}{k-1}\right)$, is convex on $(0,c)$ and concave on $(c,1)$ for some $c$. If $y=\frac 1k$ is the local minimum, then the maximum occurs at $y=0$ (recall $y\leq\frac 1k$). If $y=\frac 1k$ is the local maximum, then the maximum occurs at $y=0$ or $y=\frac 1k$.

In either case, it follows that the maximizing solution set consists of a certain number of equal values and the rest of the values being $0$.

Note: A full, rigorous solution of this problem now reduces to the following:

1. Extending of River Li's answer to $a>2$.
2. Showing $g''(y)$ satisfies the above convex-concave property.

Then, the maximizing solution set reduces to that of Dr. Wolfgang Hintze's post, where direct comparison can be used to find the maximizing set for every $a$.

Answer 3

Update Nov. 23, 2024

I have continued my experiments and I believe I have found a method to calculate the "phase"-transition points for the parameter $a$, and have - in view of the seeming complexity of the problem suprisingly - obtained an "almost" analytic solution.

The solution

The solution can be given in terms of a simple function the domain of which is determined by a set of numerically determined global transition constants.

Here I present just the solution leaving explanations for later.

Recall that defining the sum

$$s(a,n,\{p\}) = \sum_{i=1}^{n} p_{i}^{(1-p_{i})^a} \tag{1}$$

where $a>0$ is a real parameter and $p_{i}\in [0,1]$ are probabilities subject to $\sum_{i=1}^{n} p_{i}=1 $ we are looking, for given $n$ and $a$, for the maximum $s_{max}$ of $s$ over all possible sets $\{p\}=\{p_{1}, p_{2}, \cdots , p_{n}\}$.

Now for $k\ge 2$ define the set of functions

$$f(k,a)=k^{1-\left(1-\frac{1}{k}\right)^a}\tag{2}$$

then the solution is given by

$$s_{max}(n,a)=max(f(2,a), f(3,a), \cdots, f(n,a))\\ = \max \left(2^{1-\left(\frac{1}{2}\right)^{a}},3^{1-\left(\frac{2}{3}\right)^a},\cdots,n^{1-\left(\frac{n-1}{n}\right)^a}\right) \tag{3}$$

This can be made more explicit by calculating first the set of transition points $a_t(k)$ between each two adjacent functions $f(k,a)$ defined by the solution of the equation

$$f(k,a_t(k))=f(k+1, a_t(k))\tag{4}$$

This equation can be solved iteratively starting with the inial value $a_t(k) = 2k$.

The first 10 values, starting at $k=2$, are

$$a_{t}(k=2..10)=\{0.553953,2.13996,3.87677,5.72384,\\ 7.65616,9.65727,11.7156,13.8228,15.9724\}\tag{5}$$

Then we can write $s_{max}$ as this piecewise function

$$s_{max}(n,a) = \text{IF } a\in (a_t(k),a_t(k+1)) \text{ THEN} f(k,a) \tag{6}$$

Note that $s_{max}(n,a)$ is a continuous function of $a$ but it is not smooth, i.e. its derivative is not continuous but has (finite) jumps at the transition points.

Example solutions

The general behaviour of the solution can be described as follows, also illustrated by the simulations:

For $n \gt2$ and very small $a$ the maximum is $a_{max}(n,a)=f(2,a)$ corresponding to the probability set $\{p\} = \{\frac{1}{2},\frac{1}{2}, 0,\cdots,0\}$, when a grows beyond the transition point $a_t(2)$ the maximum becomes $a_{max}(n,a)=f(3,a)$ corresponding to the set $\{p\} = \{\frac{1}{3},\frac{1}{3}, \frac{1}{3}, 0, \cdots,0\}$, and so on up to $a_t(n-1)<a<a_t(n)$ where $a_{max}(n,a)=f(n-1,a)$ with $\{p\} = \{\frac{1}{n-1},\cdots,\frac{1}{n-1},0\}$ and, finally, for $a>a_t(n)$ where $a_{max}(n,a)=f(n,a)$ with $\{p\} = \{\frac{1}{n},\cdots,\frac{1}{n}\}$.

(Examples and graphs $s_{max}(a)$ to be done)

Original post

In order to get an overwiew of the solutions I have carried out some Monte-Carlo simulations and found an interesting phase transition like behaviour.

Let us rephrase the problem. Define the sum

$$s(a,n,\{p\}) = \sum_{i=1}^{n} p_{i}^{(1-p_{i})^a} \tag{1}$$

where $a>0$ is a real parameter and $p_{i}\in [0,1]$ are probabilities subject to $\sum_{i=1}^{n} p_{i}=1 $.

For given $n$ and $a$ we are looking for the maximum $s_{max}$ of $s$ over all possible sets $\{p\}=\{p_{1}, p_{2}, \cdots , p_{n}\}$.

Monte Carlo pseudo code

In order to find the maximum for given parameter $a$ we set up a Monte-Carlo simulation in the following loop

10 max = 0
20 get a set of $n$ random probabilities $\{p\}$ as defined above
30 calculate s
40 If s>max then (max = s; $\{p\}_{max} = \{p\}$)
50 goto 20

The loop is carried out a certain number $r$ of times until the maximum approaches a constant value.

Notice that we obtain not only the value of the maxiumum but - and this is even more interesting - the set of probabilities leading to that maximum.

Results

Results of a preliminary study are

For given $n= 2 .. 7$ and for $a=\{0.1, 1, 10\}$ we calculate the maximum of $s$ ad depict the set of probabilities corresponding to that value.

Actually, for the time beeing I found it more interesting to study the structure of the peobability sets than the maximum (which I shall provide soon).

All values (except for n=2) of the probabilities are approximated by simple rationals in order to exhibit the structure of the set more clearly.

We can see that, with varying $a$, there are interesting transistions in the structure of the probability sets. These remind (me) of phase transistions.

This should be studies in more depth but the available results already seem to be interesting.

$n=2$ ($r = 10^5$)
$a=0.1$ -> $\{p\} =$ {$\frac {1}{2}$,$\frac {1}{2}$}
$a=1.0$ -> $\{p\} =$ {$\frac {1}{2}$,$\frac {1}{2}$}
$a=10$ -> $\{p\} =$ {$\frac {1}{2}$,$\frac {1}{2}$}

$n=3$ ($r = 10^5$)
$a=0.1$ -> $\{p\} =$ {$\frac {1}{2}$,$\frac {1}{2}$,$0$}
$a=0.55$ -> $\{p\} =$ {$\frac {1}{2}$,$\frac {1}{2}$,$0$}
$a=0.56$ -> $\{p\} =$ {$\frac {1}{3}$,$\frac {1}{3}$,$\frac {1}{3}$}
$a=1.0$ -> $\{p\} =$ {$\frac {1}{3}$,$\frac {1}{3}$,$\frac {1}{3}$}
$a=10$ -> $\{p\} =$ {$\frac {1}{3}$,$\frac {1}{3}$,$\frac {1}{3}$}

$n=4$ ($r = 10^5$)
$a=0.1$ -> $\{p\} \simeq$ {$\frac {1}{2}$,$\frac {1}{2}$,$0$,$0$}
$a=0.55$ -> $\{p\} =$ {$\frac {1}{2}$,$\frac {1}{2}$,$0$,$0$}
$a=0.56$ -> $\{p\} =$ {$\frac {1}{3}$,$\frac {1}{3}$,$\frac {1}{3}$,$0$}
$a=1.0$ -> $\{p\} =$ {$\frac {1}{3}$,$\frac {1}{3}$,$\frac {1}{3}$,$0$}
$a=2.1$ -> $\{p\} =$ {$\frac {1}{3}$,$\frac {1}{3}$,$\frac {1}{3}$,$0$}
$a=2.2$ -> $\{p\} =$ {$\frac {1}{4}$,$\frac {1}{4}$,$\frac {1}{4}$,$\frac {1}{4}$}
$a=10$ -> $\{p\} =$ {$\frac {1}{4}$,$\frac {1}{4}$,$\frac {1}{4}$,$\frac {1}{4}$}

$n = 5$ ($r = 10^5$)
$a = 0.1$ -> $\{p\} =$ {$\frac {1}{2}$,$\frac {1}{2}$,$0$,$0$,$0$}
$a = 0.47$ -> $\{p\} =$ {$\frac {1}{2}$,$\frac {1}{2}$,$0$,$0$,$0$}
$a = 0.48$ -> $\{p\} =$ {$\frac {1}{3}$,$\frac {1}{3}$,$\frac {1}{3}$,$0$,$0$}
$a = 1.0$ -> $\{p\} =$ {$\frac {1}{3}$,$\frac {1}{3}$,$\frac {1}{3}$,$0$,$0$}
$a = 3.7$ -> $\{p\} =$ {$\frac {1}{3}$,$\frac {1}{3}$,$\frac {1}{3}$,$0$,$0$}
$a = 3.8$ -> $\{p\} =$ = {$\frac {1}{5}$,$\frac {1}{5}$,$\frac {1}{5}$,$\frac {1}{5}$,$\frac {1}{5}$}
$a = 10$ -> $\{p\} =$ = {$\frac {1}{5}$,$\frac {1}{5}$,$\frac {1}{5}$,$\frac {1}{5}$,$\frac {1}{5}$}

$n = 6$ ($r = 10^5$)
$a = 0.1$ -> $\{p\} =$ {$\frac {1}{2}$,$\frac {1}{2}$,$0$,$0$,$0$,$0$}
$a = 1.0$ -> $\{p\} =$ {$\frac {1}{3}$,$\frac {1}{3}$,$\frac {1}{3}$,$0$,$0$,$0$}
$a = 10$ -> $\{p\} =$ {$\frac {1}{6}$,$\frac {1}{6}$,$\frac {1}{6}$,$\frac {1}{6}$,$\frac {1}{6}$,$\frac {1}{6}$}}

$n = 7$ ($r = 10^5$)
$a = 0.1$ -> $\{p\} =$ {$\frac {1}{2}$,$\frac {1}{2}$,$0$,$0$,$0$,$0$,$0$}
$a = 1.0$ -> $\{p\} =$ {$\frac {1}{3}$,$\frac {1}{3}$,$\frac {1}{3}$,$0$,$0$,$0$,$0$}
$a = 10$ -> $\{p\} =$ {$\frac {1}{7}$,$\frac {1}{7}$,$\frac {1}{7}$,$\frac {1}{7}$,$\frac {1}{7}$,$\frac {1}{7}$,$\frac {1}{7}$}

Answer 4

Some thoughts 2.

In my previous answer, we dealt with the case $0 < a \le 2$. Here we deal with the case $a > 2$.

Fact 1. $f(x)$ is convex on $[0, c]$, and concave on $[c, 1]$, for some $c \in (0, 1)$.
(The proof is given at the end.)

Now we proceed. WLOG, assume that $p_1 \le p_2 \le \cdots \le p_n$. Let $$F(p_1, p_2, \cdots, p_n) := f(p_1) + f(p_2) + \cdots + f(p_n).$$ Then $F$ is maximal for $p_1 = \cdots = p_{k-1} = 0$ and $p_{k+1} = p_{k+2} = \cdots = p_n$ for some $k \in \{1, 2, \cdots, n\}$. The proof is easy. But many year ago, I knew Bin Zhao's half concave half convex theorem: https://artofproblemsolving.com/community/c6h64933. So see the proof there.

Cleary $k < n$ (when $k=n$, $(p_1, \cdots, p_n) = (0, \cdots, 0, 1)$ which is not a maximizer). Let $p_k = y$. Then $y_{k+1} = \cdots = y_n = \frac{1-y}{n-k}$. We have $$F = y^{(1-y)^a} + (n-k)\left(\frac{1-y}{n-k}\right)^{\left(1-\frac{1-y}{n-k}\right)^a}.$$

Can we proceed?

$\phantom{2}$

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Proof of Fact 1.

We have $$f''(x) = x^{(1-x)^a}(1-x)^{2a} g(x)$$ where \begin{align*} g(x) &:= \left(- \frac{a\ln x}{1-x} + \frac{1}{x}\right)^2 + \left(\frac{a^2\ln x}{(1-x)^2} - \frac{a\ln x}{(1-x)^2} - \frac{2a}{x(1-x)} - \frac{1}{x^2}\right)\frac{1}{(1-x)^a}. \end{align*}

We can prove that $g'(x) < 0$ on $(0, 1)$ (the proof is given at the end).

Note that $\lim_{x\to 0^{+}} g(x) = \infty$ and $\lim_{x\to 1^{-}} g(x) = -\infty$. Thus, there exists $c \in (0, 1)$ such that $g(x) > 0$ on $(0, c)$ and $g(x) < 0 $ on $(c, 1)$. Thus, there exists $c \in (0, 1)$ such that $f''(x) > 0$ on $(0, c)$ and $f''(x) < 0 $ on $(c, 1)$. The desired result follows.

$\phantom{2}$

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Proof of $g'(x) < 0$ on $(0, 1)$.

After some manipulations, it suffices to prove that, for all $0 < x < 1$, $$M (1 - x)^a + N < 0,$$ where ($Q = \ln x$) \begin{align*} M &:= - 2(-Qax - x + 1)[ax(Qx - x + 1) + (x-1)^2], \\ N &:= a \left( a+2 \right) \left( a-1 \right)x^3 Q \\ &\qquad +{a}^{2}{x}^{3}-{a}^{ 2}{x}^{2}+4\,a{x}^{3}-5\,a{x}^{2}-2\,{x}^{3}+ax+6\,{x}^{2}-6\,x+2. \end{align*} It is easy to prove that $M < 0$ on $(0, 1)$.

We claim that $N < 0$ for all $\frac{1}{a} \le x < 1$. Using $\ln x \le \frac{2(x-1)}{1+x}$ for all $x\in (0, 1)$, it suffices to prove that, for all $x\in [1/a, 1)$, \begin{align*} &a \left( a+2 \right) \left( a-1 \right)x^3 \cdot \frac{2(x - 1)}{1 + x} \\ &\qquad +{a}^{2}{x}^{3}-{a}^{ 2}{x}^{2}+4\,a{x}^{3}-5\,a{x}^{2}-2\,{x}^{3}+ax+6\,{x}^{2}-6\,x+2 < 0 \end{align*} which is true. The claim is proved

Thus, it suffices to prove that, for all $0 < x < 1/a$, $$M(1 - x)^a + N < 0. $$

Using Bernoulli inequality $(1+u)^r \ge 1 + ur$ for all $r \ge 1$ and $u \ge -1$, we have $(1-x)^a = (1-x)(1-x)^{a-1} \ge (1-x)[1 - x(a-1)]$. It suffices to prove that, for all $0 < x < 1/a$, $$M (1-x)[1 - x(a-1)] + N < 0, $$ or (after clearing the denominators) $$m_2 \ln^2 x + m_1 \ln x + m_0 < 0, \tag{1}$$ where \begin{align*} m_2 &:= 2a^2(1-x)x^2(1+x-ax), \\ m_1 &:= -2\,{a}^{3}{x}^{4}+4\,{a}^{3}{x}^{3}+6\,{a}^{2}{x}^{4}-{a}^{3}{x}^{2}- 12\,{a}^{2}{x}^{3}\\ &\qquad -4\,a{x}^{4}+7\,{a}^{2}{x}^{2}+6\,a{x}^{3}-6\,ax+2\, a,\\ m_0 &:= -2\,{a}^{2}{x}^{4}+6\,{a}^{2}{x}^{3}+4\,a{x}^{4}-5\,{a}^{2}{x}^{2}-12 \,a{x}^{3}\\ &\qquad -2\,{x}^{4}+{a}^{2}x+16\,a{x}^{2}+6\,{x}^{3}-9\,ax-6\,{x}^{2 }+a+2\,x. \end{align*} Clearly, we have $m_2 > 0$.

Using $-\ln x \le \frac{1 - x^2}{2x}$ for all $0 < x < 1$, to prove (1), it suffices to prove that $$m_2 (-\ln x) \cdot \frac{1 - x^2}{2x} + m_1 \ln x + m_0 < 0, $$ or $$\left(- m_2 \cdot \frac{1 - x^2}{2x} + m_1 \right)\ln x + m_0 < 0. \tag{2}$$ We can prove that $- m_2 \cdot \frac{1 - x^2}{2x} + m_1 > 0$.

Using $\ln x \le \frac{2(x-1)}{1+x}$ for all $0 < x < 1$, to prove (2), it suffices to prove that $$\left(- m_2 \cdot \frac{1 - x^2}{2x} + m_1 \right)\cdot \frac{2(x-1)}{1+x} + m_0 < 0,$$ or \begin{align*} &2\,{a}^{3}{x}^{5}-6\,{a}^{3}{x}^{4}-2\,{a}^{2}{x}^{5}+6\,{a}^{3}{x}^{3 }+10\,{a}^{2}{x}^{4}-18\,{a}^{2}{x}^{3}\\ &\qquad -4\,a{x}^{4}+17\,{a}^{2}{x}^{2} +8\,a{x}^{3}-2\,{x}^{4}-3\,{a}^{2}x+2\,{x}^{3}\\ &\qquad -5\,ax+2\,{x}^{2}+3\,a-2 \,x > 0 \end{align*} which is true. We can prove it by the Buffalo Way (BW), i.e., use the substitution $a = 2 + t, x = \frac{1}{a} \cdot \frac{1}{1+s}$ for $s, t > 0$.

We are done.