# Trying to prove an inequality (looks similar to entropy)

I'm trying to prove the following inequality (or something similar, up to a constant factor in either side of the inequality): $$k\cdot\sum_{i=1}^{k}x_{i}\cdot\ln\left(x_{i}\right)\geq\sum_{i=1}^{k}x_{i}\cdot\left(x_{i}-1\right)$$ where $$\forall i\in\left[k\right]$$, $$x_i \in\left[0,k\right]$$ (the $$x_i$$s are not necessarily natural numbers, but we can assume that they're rational if it helps), and $$\sum_{i=0}^k x_i=k$$.

I've tried plotting it for $$k=2,3$$ and ran some numerical experiments for larger $$k$$, and I'm 99% sure this inequality is correct, but I'm still struggling with the proof.

Up to some normalizing, I find the left-hand side quite similar to the entropy of a probability distribution, but I didn't manage to take advantage of this fact either. I also tried looking for inequalities that only hold on simplex-like hyperplanes, but couldn't find anything useful.

Any ideas? Thanks!

• Rewrite the RHS as $\sum_i(x_i-1)^2$ and use the first order Taylor formula at $1$ with the remainder in the Lagrange form for the function $x\mapsto x\log x$ on the LHS. That will immediately give you "something similar" with $2k$ instead of $k$ on the left. Then you can tweak this idea a bit to improve the constant. Commented Sep 2, 2021 at 9:25

Expanding a little bit on fedja's comment, it is actually convenient to consider the inequality in the equivalent form $$\sum_{i=1}^k ( x_i \ln x_i - x_i + 1) \geq c_k \sum_{i=1}^k (x_i-1)^2 ,$$ where we are interested to find $$c_k$$ such that the inequalities hold for all $$x_i \in [0,k]$$ with $$\sum_{i=1}^k x_i = k$$.
By Taylor, it is easy to see, that we have the two inequalities $$x \ln x -x +1 \geq \tfrac{1}{2} (x-1)^2 \qquad\text{ for } x \in [0,1]$$ and $$x \ln x -x +1 \leq \tfrac{1}{2} (x-1)^2 \qquad\text{ for } x \geq 1$$ Hence, we are only in trouble for $$x\in [1,k]$$. By comparing the derivatives of the function on the left hand and right hand side: $$\ln(x)$$ and $$x-1$$, we also see that the defect in the second estimate is monotone increasing in $$x$$ starting at $$x=1$$, where both are equal.
Hence, we find the optimal constant as $$c_k = \inf_{x\in [0,k]} \frac{x \ln x -x +1}{(x-1)^2} = \frac{k \ln k - k +1}{(k-1)^2}.$$ Note that $$c_k$$ is monotone decreasing in $$k$$ with $$c_1=1/2$$. In particular, it holds $$c_k \geq \frac{1}{2k}$$, but this gives not the correct asymptotics, since for $$k\geq 2$$, one can use the lower bound $$c_k \geq \frac{\ln k -1}{k} ,$$ which is asymptotic sharp for $$k\to \infty$$.