I'm trying to figure out an upper bound on $(x^2 + z^2)^p$ for an arbitrary integer power $p \geq 1$. For instance, we have the well-known bound $(x+y)^2 \leq 3x^2 + 3y^2$, but I'm not sure how to extend this.
EDIT: For my purposes, it is sufficient to assume that $x,z \geq 0$.
On
The easiest proof was given by Alan. Let me give a slightly different one. Note that $f(x)=x^p$ is a convex function on the positive reals. Thus we get $$(x+y)^p= 2^p(x/2+y/2)^p=2^p f(x/2+ y/2)\leq 2^{p-1}(f(x)+f(y))= 2^{p-1}(x^p+y^p).$$ Note that this estimate is sharp (pick $x=y$). However, it only works for $p\geq 1$.
There's a simple trick to get the bound you want. Suppose $0\leq a\leq b$, then $$ (a+b)^p\leq (2b)^p=2^pb^p\leq 2^p(a^p+b^p). $$ By simmetry we get the same inequality for $0\leq b\leq a$. So in any case we have $$(a+b)^p\leq 2^p(a^p+b^p).$$
Note that $p$ can be any nonnegative real number.