I want to show that the function $(x+y)^{c}$, with $x,y\geq0$ and $c\in(0,1)$. From the definition of concavity, it amounts to show
\begin{align} (\lambda x_{0} + \lambda y_{0} + (1-\lambda) x_{1} + (1-\lambda) y_{1})^{c} \geq \lambda(x_{0}+y_{0})^{c} + (1-\lambda)(x_{1}+y_{1})^{c} \end{align}
and the Hessian is given by
$$ H = c(c-1)(x+y)^{c-2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} $$
However, I seem to be missing something.
Let $f(x,y)=(x+y)^c$, further let $v_0=(x_0,y_0)$, $v_1=(x_1,y_1)$ be such that $x_0+y_0$, $x_1+y_1\ge 0$, $\lambda\in(0,1)$ and $v_\lambda=(x_\lambda,y_\lambda)=\lambda v_1+(1-\lambda)v_0$. If one sum is $0$, say $x_0+y_0=0$, we have $f(v_\lambda)=\lambda^cf(v_1)\ge\lambda f(v_1)+(1-\lambda)f(v_0)$ using $f(v_0)=0$ and $c\in(0,1)$. Otherwise let $g:[0,1]\rightarrow\mathbb R$, $t\mapsto f(v_t)$. For $t\in[0,1]$ we have $x_t+y_t=t(x_1+y_1)+(1-t)(x_0+y_0)>0$, so $g'(t)=c(x_t+y_t)^{c-1}\Delta$, where $\Delta=x_1+y_1 -x_0-y_0$, and $g''(t)=c(c-1)(x_t+y_t)^{c-2}\Delta^2$. Using $c\in(0,1)$ we have $g''(t)<0$ for all $t\in[0,1]$ (take one sided derivatives or slightly extend $g$ in both directions). Hence, for any $0\le t\le s\le 1$ we have $g'(s)-g'(t)=\int_t^sg''(u)\mathrm du\le0$ with equality if and only if $t=s$. This shows that the derivative is strictly decreasing. Using the mean value theorem with $g(1)-g(0)=\int_0^1g'(t)\mathrm dt$ we obtain $t^*\in[0,1]$ with $g'(t^*)=g(1)-g(0)$. In particular, for $h:[0,1]\rightarrow\mathbb R$, $t\mapsto g'(t^*)t+g(0)$ we have $h(1)=g(1)$. With $g'(0)=\int_0^1g'(0)\mathrm dt>\int_0^1g'(t)\mathrm dt=g'(t^*)$, and analogously $g'(1)<g'(t^*)$, we have $t^*\in(0,1)$, so $g(t)=g(0)+\int_0^{t}g'(u)\mathrm du>g(0)+g'(t^*)t=h(t)$ for $t\in(0,t^*]$. Assume there exists $t\in(t^*,1)$ with $g(t)\le h(t)$, then $g(1)=g(t)+\int_t^1g'(u)\mathrm du<h(t)+g'(t^*)(1-t)=h(1)=g(1)$ is a contradiction, so $g(t)>h(t)$ for all $t\in (0,1)$, in particular $f(v_\lambda)=g(\lambda)>h(\lambda)=f(v_0)+g'(t^*)\lambda=\lambda f(v_1)+(1-\lambda) f(v_0)$, since $g'(t^*)=g(1)-g(0)=f(v_1)-f(v_0)$. This shows that $f$ is convex on the maximal domain given by $x+y\ge 0$, and strictly convex on the interior.
Although we avoided the Hessian alltogether, we have $g''<0$ because $H$ is negative definite. However, I prefer to consider $g$ (instead of $f$) from the very beginning, since this reduces the multidimensional case to the one-dimensional case, without affecting the arguments. Also, should this not be clear, any function for which the Hessian exists everywhere and is negative semidefinite, is concave (cf. the remark here that clarifies the relation of concave and convex functions, and the remark here that relates convexity to the Hessian).