Take any $r\in(-1,0)$, any vector $(p_i)_{i=1}^n$ with $p_i>0$ for all $i$, and any vector $(h_i)_{i=1}^n\in\mathbb R^n$. For all real $t$ close enough to $0$, let $$g(t):=M(p+th).$$ Then $$g''(0)=\Big((1+r)\sum_{i=1}^n p_i^{r-1}h_i^2\,\sum_{i=1}^n p_i^{1+r} +\frac{1-r^2}{r^2}\,\Big(\sum_{i=1}^n p_i^r h_i\Big)^2\Big) \Big(\sum_{i=1}^n p_i^{1+r}\Big)^{1/r-2},$$ which is manifestly $\ge0$.

It follows that $M$ is indeed convex (actually, convex on the entire positive orthant of $\mathbb R^n$).

Take any $r\in(-1,0)$, any vector $(p_i)_{i=1}^n$ with $p_i>0$ for all $i$, and any vector $(h_i)_{i=1}^n\in\mathbb R^n$. For all real $t$ close enough to $0$, let $$g(t):=M(p+th).$$ Then $$g''(0)=(1+r)\Big(\sum_{i=1}^n p_i^{r-1}h_i^2\,\sum_{i=1}^n p_i^{1+r} +\frac{1-r^2}{r^2}\,\Big(\sum_{i=1}^n p_i^r h_i\Big)^2\Big) \Big(\sum_{i=1}^n p_i^{1+r}\Big)^{1/r-2},$$ which is manifestly $\ge0$.

It follows that $M$ is indeed convex (actually, convex on the entire positive orthant of $\mathbb R^n$).

Take any $r\in(-1,0)$, any vector $(p_i)_{i=1}^n$ with $p_i>0$ for all $i$, and any vector $(h_i)_{i=1}^n\in\mathbb R^n$. For all real $t$ close enough to $0$, let $$g(t):=M(p+th).$$ Then $$g''(0)=(1+r)\sum_{i=1}^n p_i^{r-1}h_i^2\,\sum_{i=1}^n p_i^{1+r} +\frac{1-r^2}{r^2}\,\Big(\sum_{i=1}^n p_i^r h_i\Big)^2,$$ which is manifestly $\ge0$.

It follows that $M$ is indeed convex (actually, convex on the entire positive orthant of $\mathbb R^n$).

Take any $r\in(-1,0)$, any vector $(p_i)_{i=1}^n$ with $p_i>0$ for all $i$, and any vector $(h_i)_{i=1}^n\in\mathbb R^n$. For all real $t$ close enough to $0$, let $$g(t):=M(p+th).$$ Then $$g''(0)=\Big((1+r)\sum_{i=1}^n p_i^{r-1}h_i^2\,\sum_{i=1}^n p_i^{1+r} +\frac{1-r^2}{r^2}\,\Big(\sum_{i=1}^n p_i^r h_i\Big)^2\Big) \Big(\sum_{i=1}^n p_i^{1+r}\Big)^{1/r-2},$$ which is manifestly $\ge0$.

It follows that $M$ is indeed convex (actually, convex on the entire positive orthant of $\mathbb R^n$).