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?
 A: Observe that $f(x)=x^p$ for $p\in\mathbb N$ is a convex function over $\mathbb R$. Therefore we can use Jensen inequality and we get:
$$
\frac{x^p+y^p}2\geq (\frac{x+y}2)^p
$$
which proves the following inequality for all $x,y$:
$$
x+y\le(2^{p-1}x^p+2^{p-1}y^p)^{1/p}
$$
However this is more tricky for the sequence of inequalities you wrote down. Suppose that $1<q<p$. define the function $f(x)=x^{\frac pq}$. The function is convex over its domain which is not necessarily $\mathbb R$. We get back to the question of what is the domain of this function. For the moment see that the function is convex over its domain and hence we can use Jensen again to see that:
$$
\frac{x^p+y^p}2=\frac{(x^q)^{\frac pq}+(y^q)^{\frac pq}}2\geq \left(\frac{x^q+y^q}2\right)^{\frac pq}
$$
which means that:
$$
(2^{q-1}x^q+2^{q-1}y^q)^{1/q} \leq (2^{p-1}x^p+2^{p-1}y^p)^{1/p}.
$$
The domain of function however is not always $\mathbb R$. For instance $f(x)=x^{\frac 32}$ is defined over $\mathbb R^+$. 
In general one can say that for odd $q$ or even $p$ the domain is still $\mathbb R$ and the previous inequality is true for $x,y\in\mathbb R$. 
However for even $q$ and odd $p$, this is true only on  $\mathbb R^+$. for instance choose $x=-y$, $p=3$ and $q=2$ which shows a counter example.
$$
x+y\le(2x^2+2y^2)^{1/2}\le...\le(2^{p-1}x^p+2^{p-1}y^p)^{1/p}
$$
A: 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.
