Here is my homework from convex optimization.
Let $f:(0,\infty)^{n}\rightarrow (0,\infty)$ be given by $f(x)=\left(\sum\limits_{j=1}^{n}\alpha_{j}x_{j}^{p}\right)^{\frac{1}{p}}$, where $\alpha_{j}>0$, $\sum\limits_{j=1}^{n}\alpha_{j}=1,p\in(\infty,1)\backslash\{0\}$. Show, using a gradient monotonicity of function -$f$, that $f$ is concave.
I know that in order to show the concavity of $f$ I need to show a convexity of $g:=-f$ and in terms of gradient monotonicity it means that I need to show that $(\nabla g(y) - \nabla g(x))\circ(y-x)\geq 0$ for any $x,y\in\mathbb{R}^{n}$. So I've computed gradient of $g$ and I've got $\nabla g(y)=\left[-\alpha_{1}x_{1}^{p-1}\left(\sum\limits_{j=1}^{n}\alpha_{j}x_{j}^{p}\right)^{\frac{1}{p}-1},...,-\alpha_{n}x_{n}^{p-1}\left(\sum\limits_{j=1}^{n}\alpha_{j}x_{j}^{p}\right)^{\frac{1}{p}-1}\right]$. Then I started to compute $(\nabla g(y) - \nabla g(x))\circ(y-x)$ and after some manipulations I've obtained that
$$(\nabla g(y) - \nabla g(x))\circ(y-x)=\sum\limits_{i=1}^{n}{\frac{y_{i}}{x_{i}}}\alpha_{i}x_{i}^{p}\left(\sum\limits_{j=1}^{n}\alpha_{j}x_{j}^{p}\right)^{\frac{1}{p}-1}-\left(\sum\limits_{j=1}^{n}\alpha_{j}x_{j}^{p}\right)^{\frac{1}{p}}+\sum\limits_{i=1}^{n}{\frac{x_{i}}{y_{i}}}\alpha_{i}y_{i}^{p}\left(\sum\limits_{j=1}^{n}\alpha_{j}y_{j}^{p}\right)^{\frac{1}{p}-1}-\left(\sum\limits_{j=1}^{n}\alpha_{j}y_{j}^{p}\right)^{\frac{1}{p}}$$ but this is a moment when I don't know how to proceed. I've tried to find something on Internet but the only results showing this function is concave (it turned out that this is so-called CES function) were based on so-called quasi-concavity, not gradient monotonicity. Any hints would be greatly appreciated.