Prove that $(x^{\frac{1}{p}}+y^{\frac{1}{p}})^p$ is concave. Let $g(x,y) = (x^{\frac{1}{p}}+y^{\frac{1}{p}})^p$  $ x>0,y>0, p>1$
Need to show that $g(x,y)$ is concave. 
I tried it by finding out the partial derivative matrix and showing that it is negative semi-definite. However, this method is too cumbersome.
Can anyone come up with any other elegant method for proving it?
Any hints will be appreciated.
 A: Let $G_a(z) = z^a$, and $F(x, y) = G_{1/p}(x) + G_{1/p}(y)$, then $g(x,y) = G_p(F(x,y))$.
For $G_a(z) = z^a$, we have
$$
G'_a(z) = a \, z^{a-1}.
\tag{1}
$$
and
$$
G''_a(z) = a \, (a-1) \, z^{a-2}.
\tag{2}
$$
Now for $g(x,y)$. Since we have two variables, we need to show for any
\begin{align}
x' &= x + c \, t \\
y' &= y + s \, t,
\end{align}
where $c$ and $s$ are two real numbers for the two directional cosines,
we have
$$
\left. \frac{ \partial^2 g(x + c\,t, y + s\,t) }{
 \partial t^2 } \right|_{t=0}
< 0.
$$
But $g(x,y) = G_p(F(x,y))$, by the chain rule,
\begin{align}
\frac{\partial g(x, y) } {\partial t}
&=
G'_p(F) \left[
\frac{\partial F(x,y)}{\partial x} \, c
+
\frac{\partial F(x,y)}{\partial y} \, s
\right] \\
&=
G'_p(F) \left[
\frac{\partial G_{1/p}(x)}{\partial x} \, c
+
\frac{\partial G_{1/p}(y)}{\partial y} \, s
\right] \\
&=
G'_p(F) \left[
G'_{1/p}(x) \, c
+
G'_{1/p}(y) \, s
\right].
\end{align}
and
\begin{align}
\frac{\partial^2 g(x, y) } {\partial t^2}
&=
G''_p(F) \left[
G'_{1/p}(x) \, c
+
G'_{1/p}(y) \, s \right]^2
+
G'_p(F) \left[
G''_{1/p}(x) \, c^2
+
G''_{1/p}(y) \, s^2 \right]
 \\
&=
\left(\frac{1}{p}-1\right)
F^{p-2}\left[
 \left(
   x^{1/p} + y^{1/p}
 \right)
 \left(
   c^2 \, x^{1/p - 2} + s^2 \, y^{1/p - 2}
 \right)
 -
\left(
  c \, x^{1/p-1} + s \, y^{1/p-1}
\right)^2
\right]
.
\end{align}
The term in the square brackets is nonnegative by the Cauchy-Schwarz inequality, and $1/p - 1 <0$.  So $g(x,y)$ is concave.
