I was given an open problem as a birthday present recently. While I can probably handle spoilers at this point, what I really want are literature and other references. Also acceptable would be suggestions for approaches. Ideally, search terms for the web are most welcome. (It is my lack of imagination for search terms for this problem that is most keeping me from doing preliminary research on it.)


**Question:** Where and how can I find out more about the Problem ($C\_p \lt 2$?) below?


Problem: Let $n$ be a an integer parameter sufficiently large. Let $x$ and $y$ be vectors in $n$-dimensional real space. ( Do complex if it makes it easier. ) We will be using a family of $L\_p$ norms $|z|$ for $p$ real and greater than $1$ ( $p$th root of the sum of the $p$th powers of the $n$ coordinates of the vector $z$), although other norms are also of interest. Fixing $p$ and its associated norm, there is a value $C\_p$ ( which may be independent of $n$ and all other parameters ) such that for all $x$ and $y$ with $|x|=|y|=|x-y|=1$, one has $|x+y|\leq C\_p$. The triangle inequality gives $C\_p \leq 2$, and for $p=2$ we have $C\_2 = \sqrt{3}$. Is it true that $C\_p \lt 2$? (Can we also hope that lim sup over all $p \gt 1$ of $C\_p$ is also less than $2$?)


$C\_p$ to me represents the length of the long diagonal of a rhombus (diamond) with short diagonal and sides length $1$. Please retag this post as appropriate.


Gerhard "Still Enjoys His Birthday Presents" Paseman, 2015.11.20



Looking for (information about) long diamonds
Here is a simple short proof of one of the first main questions.



> 
> **Claim.** $C\_p < 2$.
> 
> 
> 


*Proof.* Let $C\_p$ be above. We look at two cases: (i) $1 < p < 2$, and (ii) $p > 2$.


Case (i): From [this short note](https://web.math.princeton.edu/~naor/homepage%20files/inequality.pdf) we know that
\begin{equation\*}
\|x+y\|\_p^2 \le 2(\|x\|\_p^2 + \|y\|\_p^2) + (1-p)\|x-y\|\_p^2.
\end{equation\*}
Using the hypothesis, $\|x\|\_p=\|y\|\_p=\|x-y\|\_p=1$, we obtain
\begin{equation\*}
 \|x+y\|\_p^2 \le 5 - p\quad\implies C\_p < 2.
\end{equation\*}


Case (ii): [From Hanner's inequalities](https://en.wikipedia.org/wiki/Hanner's_inequalities) since $p > 2$ we know that 
\begin{equation\*}
\|x+y\|\_p^p + \|x-y\|\_p^p \le (\|x\|\_p+\|y\|\_p)^p + | \|x\|\_p - \|y\|\_p |^p.
\end{equation\*}
Using the hypothesis, we obtain
\begin{equation\*}
 \|x+y\|\_p \le (2^p - 1)^{1/p},\quad \implies C\_p < 2.
\end{equation\*}


The only case we skipped here is $p=2$, but in that case $C\_2=\sqrt{3}$ is already known to the OP.


