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Gibbs' inequality is equivalent to:

\begin{equation} \sum_{i} \ln q_i^{p_i}-\ln p_i^{p_i} \leq 0 \end{equation}

where $p_i,q_i \in [0,1]$ and $\sum_i p_i = \sum_i q_i=1$.

Now, a friend of mine suggested that:

\begin{equation} \sum_{i} \ln q_i^{p_i}-\ln p_i^{p_i} \leq 0 \implies \sum_{i} q_i^{p_i}-p_i^{p_i} \leq 0 \end{equation}

Right now I doubt this is true but I can't think of a counter-example.

Update: I have found experimental evidence for my friends' conjecture by running the following Python code:

import numpy as np

count = 0

for i in range(10000):
    # randomly create distributions:
    P, Q = np.random.rand(10), np.random.rand(10)
    p, q = P/np.sum(P), Q/np.sum(Q)
    
    M = np.sum([p[i]**p[i] for i in range(10)])
    m = np.sum([q[i]**p[i] for i in range(10)])
    
    if m <= M:
        count+=1

The inequality was satisfied every single time I ran this script.

Gibbs' inequality is equivalent to:

\begin{equation} \sum_{i} \ln q_i^{p_i}-\ln p_i^{p_i} \leq 0 \end{equation}

where $p_i,q_i \in [0,1]$ and $\sum_i p_i = \sum_i q_i=1$.

Now, a friend of mine suggested that assuming $p_i,q_i \in [0,1]$ and $\sum_i p_i = \sum_i q_i=1$, Gibbs' inequality implies:

\begin{equation} \sum_{i} q_i^{p_i}-p_i^{p_i} \leq 0 \end{equation}

Right now I doubt this is true but I can't think of a counter-example.

Update: I have found experimental evidence for my friends' conjecture by running the following Python code:

import numpy as np

count = 0

for i in range(10000):
    # randomly create distributions:
    P, Q = np.random.rand(10), np.random.rand(10)
    p, q = P/np.sum(P), Q/np.sum(Q)
    
    M = np.sum([p[i]**p[i] for i in range(10)])
    m = np.sum([q[i]**p[i] for i in range(10)])
    
    if m <= M:
        count+=1

The inequality was satisfied every single time I ran this script.

Gibbs' inequality is equivalent to:

\begin{equation} \sum_{i} \ln q_i^{p_i}-\ln p_i^{p_i} \leq 0 \end{equation}

where $p_i,q_i \in [0,1]$ and $\sum_i p_i = \sum_i q_i=1$.

Now, a friend of mine suggested that:

\begin{equation} \sum_{i} \ln q_i^{p_i}-\ln p_i^{p_i} \leq 0 \implies \sum_{i} q_i^{p_i}-p_i^{p_i} \leq 0 \end{equation}

Right now I doubt this is true but I can't think of a counter-example.

Gibbs' inequality is equivalent to:

\begin{equation} \sum_{i} \ln q_i^{p_i}-\ln p_i^{p_i} \leq 0 \end{equation}

where $p_i,q_i \in [0,1]$ and $\sum_i p_i = \sum_i q_i=1$.

Now, a friend of mine suggested that:

\begin{equation} \sum_{i} \ln q_i^{p_i}-\ln p_i^{p_i} \leq 0 \implies \sum_{i} q_i^{p_i}-p_i^{p_i} \leq 0 \end{equation}

Right now I doubt this is true but I can't think of a counter-example.

Update: I have found experimental evidence for my friends' conjecture by running the following Python code:

import numpy as np

count = 0

for i in range(10000):
    # randomly create distributions:
    P, Q = np.random.rand(10), np.random.rand(10)
    p, q = P/np.sum(P), Q/np.sum(Q)
    
    M = np.sum([p[i]**p[i] for i in range(10)])
    m = np.sum([q[i]**p[i] for i in range(10)])
    
    if m <= M:
        count+=1

The inequality was satisfied every single time I ran this script.

Gibbs' inequality is equivalent to:

\begin{equation} \sum_{i} \ln q_i^{p_i}-\ln p_i^{p_i} \geq 0 \end{equation}

where $p_i,q_i \in [0,1]$ and $\sum_i p_i = \sum_i q_i=1$.

Now, a friend of mine suggested that:

\begin{equation} \sum_{i} \ln q_i^{p_i}-\ln p_i^{p_i} \geq 0 \implies \sum_{i} q_i^{p_i}-p_i^{p_i} \geq 0 \end{equation}

Right now I doubt this is true but I can't think of a counter-example.

Gibbs' inequality is equivalent to:

\begin{equation} \sum_{i} \ln q_i^{p_i}-\ln p_i^{p_i} \leq 0 \end{equation}

where $p_i,q_i \in [0,1]$ and $\sum_i p_i = \sum_i q_i=1$.

Now, a friend of mine suggested that:

\begin{equation} \sum_{i} \ln q_i^{p_i}-\ln p_i^{p_i} \leq 0 \implies \sum_{i} q_i^{p_i}-p_i^{p_i} \leq 0 \end{equation}

Right now I doubt this is true but I can't think of a counter-example.