Is i true that $x+y\le(2x^2+2y^2)^{1/2}\le...\le(2^{p-1}x^p+2^{p-1}y^p)^{1/p}$ Is it true that,
$$
x+y\le(2x^2+2y^2)^{1/2}\le...\le(2^{p-1}x^p+2^{p-1}y^p)^{1/p}
$$
How to prove or disprove it?

The first link is OK. $x+y \le |x+y|$, and then note we have
$$|x+y| \le (2x^2+2y^2)^{1/2} \tag{1}$$
iff $(x-y)^2 \ge 0$, after squaring both (nonnegative) sides of $(1),$ rearranging and factoring. So this gives the first inequality of the list
$x+y \le (2x^2+2y^2)^{1/2}.$
The second inequality (if I see the pattern) would be
$$(2x^2+2y^2)^{1/2} \le (4x^3+4y^3)^{1/3},$$
which fails if $x=y=-1$, so maybe you should restrict to $x,y>0$ for a possible string of inequalities.
