convexity proof of a function including ln and sums $$f(x_1,\dots,x_n)=\sum\limits_{i=1}^nx_i\ln x_i-\left(\sum\limits_{i=1}^nx_i\right)\ln\left(\sum\limits_{i=1}^nx_i\right)\rightarrow R_{++}^n$$
How can I prove this is convex on $R_{++}^n$? I tried using the Hessian and couldn't prove it. There is a solution using the gradient and Jensen but very long and complicated.
 A: Taking the Hessian gives
$$
\frac{\partial^2}{\partial x_j\partial x_k}f(x)
=\overbrace{\begin{bmatrix}
\frac1{x_1}&0&0&\cdots&0\\
0&\frac1{x_2}&0&\cdots&0\\
0&0&\frac1{x_3}&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\cdots&\frac1{x_n}
\end{bmatrix}}^{\small\displaystyle\frac{\partial^2}{\partial x_j\partial x_k}\sum_{i=1}^nx_i\log(x_i)}
-\overbrace{\frac1{\sum\limits_{i=1}^nx_i}
\begin{bmatrix}
1&1&1&\cdots&1\\
1&1&1&\cdots&1\\
1&1&1&\cdots&1\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&1&1&\cdots&1
\end{bmatrix}\vphantom{\begin{bmatrix}\frac1{x_1}\\\frac1{x_1}\\\frac1{x_1}\\\vdots\\\frac1{x_1}\end{bmatrix}}}^{\small\displaystyle\frac{\partial^2}{\partial x_j\partial x_k}\left(\sum_{i=1}^nx_i\right)\log\left(\sum_{i=1}^nx_i\right)}
$$
Thus, by Hölder's Inequality,
$$
\begin{align}
u^T\frac{\partial^2}{\partial x_j\partial x_k}f(x)u
&=\sum_{i=1}^n\frac{u_i^2}{x_i}
-\frac{\left(\sum\limits_{i=1}^nu_i\right)^2}{\sum\limits_{i=1}^nx_i}\\
&=\frac1{\sum\limits_{i=1}^nx_i}\left(\sum_{i=1}^nx_i\sum_{i=1}^n\frac{u_i^2}{x_i}-\left(\sum\limits_{i=1}^nu_i\right)^2\right)\\
&\ge0
\end{align}
$$
The Hessian Matrix is positive semi-definite, so $f$ is convex.
