How to prove this inequality $\sum_{i=1}^{n}\left(x^k_{i}\ln{x_{i}}\ln{\frac{x_{i}}{n}}\right)\le 0$ Let $x_{i}\ge 0$ for $i\in\{1,2,\cdots,n\}$ and $x_{1}+x_{2}+\cdots+x_{n}=n$ for $n\ge 3$
  
  Show that for all strictly positive integers $k\ge2$ the following inequality holds :
  $$\sum_{i=1}^{n}x^k_{i}\ln{x_{i}}\ln{\dfrac{x_{i}}{n}}\le 0$$


We consider
$$f(x)=x^k\ln{x}\ln{\dfrac{x}{n}}$$
Then
$$f'(x)=kx^{k-1}\ln{x}\cdot\ln{\dfrac{x}{n}}+x^{k-1}\ln{\dfrac{x}{n}}+\dfrac{x^{k-1}}{n}\ln{x}$$
$$\Longrightarrow 
f''(x)=k(k-1)x^{k-2}\cdot\ln{x}\cdot\ln{\dfrac{x}{n}}+kx^{k-2}\ln{\dfrac{x}{n}}+\dfrac{kx^{k-2}}{n}\ln{x}+(k-1)x^{k-2}\ln{\dfrac{x}{n}}+\dfrac{x^{k-2}}{n}+\dfrac{k-1}{n}x^{k-2}\ln{x}+\dfrac{x^{k-2}}{n}.$$
Unfortunatly I can't know  the sign of $f''(x)$ because I want to use Jensen's Inequality to prove it.

So how can we prove this inequality ?
Firstly, you diffrentiated it wrongly:
$$f(x)=x^k\ln{x}\ln{\dfrac{x}{n}}$$
$$f'(x)=kx^{k-1}\ln{x}\ln{\dfrac{x}{n}}+x^{k-1}\ln{\dfrac{x}{n}}+x^{k-1}\ln{x}$$
$$f'(x)=x^{k-1}\left(k\ln x\ln\frac xn+\ln \frac xn+\ln x\right)$$
$$f''(x)=(k-1)x^{k-2}\left(k\ln x\ln\frac xn+\ln \frac xn+\ln x\right)+x^{k-1}\left(\frac  kx\ln\frac xn+\frac kx\ln x+\frac1x+\frac1x\right)$$
$$f''(x)=(k-1)x^{k-2}\left(k\ln x\ln\frac xn+\ln \frac xn+\ln x\right)+x^{k-2}\left(k\ln\frac xn+k\ln x+2\right)$$
$$f''(x)=x^{k-2}\left(k(k-1)\ln x\ln\frac xn+(k-1)\ln\frac xn+(k-1)\ln x+k\ln\frac xn+k\ln x+2\right)$$
$$f''(x)=x^{k-2}\left(k(k-1)\ln x\ln\frac xn+(2k-1)\ln\frac {x^2}n+2\right)$$



As $k&gt;2$, $x^{k-2}&gt;0$ Also $x&lt;n$.See this graph:
I think you need to look at  $\ln\frac{x_i}{n}$ in your equation. Since, $$x_1 + x_2 + \dots + x_n = n$$
therefore $$x_i &lt; n$$
and notice that $$\ln\frac{x_i}{n} = \ln(x_i) - \ln(n) \implies \ln\frac{x_i}{n} &lt; 0$$
hence the expression $${x_i}^k\ln{x_i}\ln\frac{x_i}{n} &lt; 0$$
as ${x_i}^k$ and $\ln{x_i}$ are going to be positive. Hence the sum of all negative numbers is going to be less than zero.