TITLE: Prove this inequality $f(x_{1})+f(x_{2})+\cdots+f(x_{n})\ge\frac{n^{q+1}}{n^{q-p}-1}$
QUESTION [12 upvotes]: let $x_{i}>0,i=1,2,\cdots,n$,and such $x_{1}+x_{2}+\cdots+x_{n}=1$,Define function
  $$f(x)=\dfrac{1}{x^p-x^{q}},p>0,q\ge 1,-1<p-q<0,p,q\in \mathbb{R}$$
show that
  $$f(x_{1})+f(x_{2})+\cdots+f(x_{n})\ge\dfrac{n^{q+1}}{n^{q-p}-1}$$
  My approach is the following:

Since
$$f'(x)=-\dfrac{px^{p-1}-qx^{q-1}}{(x^p-x^q)^2},$$ Unfortunatly I can't know the sign of $f′′(x)$ because I want to use Jensen's inequality to prove it.

REPLY [2 votes]: Take $g(\textbf{x})=f(x_1)+f(x_2)+...+f(x_n)$ where $\textbf{x}=(x_1,x_2,...,x_n)$. Then a potential way toward the solution of this problem is through optimization using Lagrangian method. One could consider the following problem
$$\min_{\textbf{x}}g(\textbf{x})\hspace{0.5cm}\text{subject to}\hspace{0.5cm} \textbf{x}\cdot \textbf{1}^{T}=1$$
 where $\textbf{1}^T=\underbrace{(1,1,...,1)^T}_{n-times}$. First order conditions imply a set of simultaneous $n$ equations given by $$\partial g/\partial x_i=\lambda$$
for $i=1,2,...,n$ and $\lambda$ is a real number otherwise known as the Lagrangian multiplier. By the definition of $g(\textbf{x})$ we must have that
$$\partial g/\partial x_i\equiv f'(x_i)=-\frac{px_i^{p-1}-qx_i^{q-1}}{(x_i^p-x_i^q)^2}=\lambda$$
for all $i=1,2,...,n$. By the symmetry of this system of equations it follows that the extreme value is reached when $x_1=x_2=...=x_n$ which under the constraint yields $x_i=1/n$ for all $i=1,2,...,n$. Now to verify that at $\textbf{x}_0=(1/n, 1/n,..., 1/n)$ we have a minimum or maximum we need to check the Hessian matrix $\textbf{H}(g)$ given by second partial derivatives of the objective function $g(\textbf{x})$ and of the constraint evaluated at $\textbf{x}_0$ (you could refer to any text book in calculus, analysis or linear algebra for more details about Hessian matrices). Your problem then reduces to verifying the positive semi-definiteness of the Hessian matrix i.e. $\textbf{u}\cdot\textbf{H}\cdot\textbf{u}^T\geq 0$ for all $\textbf{u}\in\mathbb{R}^n$. If so than at $\textbf{x}_0$ we have a local minimum so 
$$g(\textbf{x})\geq g(\textbf{x}_0)=n\cdot f(1/n)=n\cdot\frac{1}{n^{-p}-n^{-q}}=\frac{n^{q+1}}{n^{q-p}-1}$$