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!

The function
$$
 f(x) = k x \ln(x) - x(x-1)
$$
has the second derivative
$$
 f''(x) = \frac k x - 2
$$
so that it is convex on the interval $[0, k/2]$. If all $x_i$ are in the interval $[0, k/2]$ then Jensen's inequality can be applied, so that
$$
 \sum_{i=1}^k f(x_i) \ge k f\left( \frac 1k  \sum_{i=1}^k x_i\right) = k f(1) = 0 \, ,
$$
which is the desired estimate.
It remains to investigate the case where $x_i > k/2$ for one $i$, say $x_k > k/2$. Then
$$
\sum_{i=1}^k f(x_i)  = \sum_{i=1}^{k-1} f(x_i) + f(x_k)
\ge (k-1) f\left( \frac 1{k-1}  \sum_{i=1}^{k-1} x_i\right) + f(x_k) \\
= (k-1) f\left( \frac {k-x_k}{k-1} \right) + f(x_k) \, .
$$
Therefore we define
$$
 g(x) = (k-1) f\left( \frac {k-x}{k-1} \right) + f(x) \, .
$$
An elementary calculation shows that $g(1) =g'(1) = 0$, and
$$
 g''(x) = \frac{2k(x^2-kx+\frac{k^2-k}{2})}{(k-1)x(k-x)} \ge 0 \, .
$$
It follows that $g(x) \ge 0$ on $[0, k]$, and that completes the proof.
