Fix $M \geq 2$. What is the smallest number $\tau = \tau(M)$ such that $$\sum\_{a,b,c,d =1\\ a + b = c+ d}^M (x\_a x\_b x\_c x\_d)^{\tau/4} \leq 1,$$ for all $x , \ldots , x\_M \in \mathbb{R}\_{\geq 0} $ satisfying $$\sum\_{j=1}^M x\_j = 1?$$ Clearly $\tau \leq 4$ and for $M = 2$, we have that $\tau = \log\_2 6$. 



A sum over a hyperplane in $\mathbb{Z}^4$
Fixed the gap. Now the argument should be complete. Feel free to ask questions if something is unclear.


The problem is equivalent to showing that the $L^4$ norm of a trigonometric polynomial $P(z)=\sum\_{k=1}^My\_kz^k$ on the unit circle $\mathbb T$ with the Haar measure $\mu$ can be bounded (with constant $1$) by the $\ell^{p}$ norm of its coefficient sequence with $p=\frac43+\frac c{\log M}$ ($y\_k^p=x\_k$). We assume that the latter norm is $1$. Note that it implies that the $\ell^{4/3}$-norm of the coefficient sequence is then at most $a(c)$ where $a(c)\to 1$ as $c\to 0+$.


*Case 1:* There exists a coefficient close to $1$. Assume that it is $y\_j=1-\varepsilon$. Then
$$
\int|P|^4\,d\mu=\int |Pz^{-j}|^4\,d\mu=\int [(1-\varepsilon)+Q]^2[(1-\varepsilon)+\bar Q]^2\,d\mu
$$
where $Q$ is a mean $0$ trigonometric polynomial with the $\ell^p$ norm of its coefficient sequence at most $[1-(1-\varepsilon)^p]^{1/p}\le 2\varepsilon^{1/p}$, so the $\ell^{4/3}$, $\ell^{3/2}$, and $\ell^2$ norms of the coefficient sequence of $Q$ are bounded by $3\varepsilon^{1/p}$, say, provided that $c>0$ is small enough. Opening the parentheses, integrating, and using Hausdorff-Young, we conclude that our integral is bounded by 
$$
(1-\varepsilon)^4+C[\varepsilon^{2/p}+\varepsilon^{3/p}+\varepsilon^{4/p}]
$$
with some absolute $C>0$.
Since all powers in brackets are strictly bigger than $1$, we stay below $1$ as long as $0\le\varepsilon\le\varepsilon\_0$ for some $\varepsilon\_0>0$.


*Case 2:* The coefficients are separated from $1$ by some $\varepsilon\_0>0$. We shall show that in this case the sum on the left with $\tau=3$ (in the original notation) is bounded by $\gamma=\gamma(\varepsilon\_0)<1$. Since it has about $M^3$ terms, the desired improvement will follow immediately just by the Holder inequality.


Note first of all that $(DCBA)^{1/4}\le \frac 14(D+C+B+A)$ and that if the ratio $B/A$, say, is outside the interval $I=[1-\varepsilon\_1,1+\varepsilon\_1]$ with some $\varepsilon\_1>0$, then we can introduce a factor $\Gamma=\Gamma(\varepsilon\_1)<1$ on the right hand side.


Now we can write 
$$
(x\_ax\_bx\_cx\_d)^{3/4}\le\frac 14[x\_ax\_bx\_c+x\_ax\_bx\_d+x\_ax\_cx\_d+x\_bx\_cx\_d]
$$
Moreover, if $x\_a/x\_b\notin I$, we can introduce the factor $\Gamma$ on the right hand side. 


Now suppose that for all $a$, we have $\sum\_{b:x\_a/x\_b\notin I}x\_b>\delta>0$. Then we get an immediate improvement 
$$
\sum\_{a,b:x\_a/x\_b\in I}x\_ax\_b+\Gamma\sum\_{a,b:x\_a/x\_b\notin I}x\_ax\_b\le 1-(1-\Gamma)\delta\,.
$$
Otherwise the main part of $x\_k$ (in terms of the sum) lies in a short (on the logarithmic scale) interval, which means that, provided $\delta>0$ is small enough, we should get the sum close to $1$ from that part alone if our claim does not hold. This part cannot be a singleton (otherwise we are in Case 1) Thus, it remains to show that for any finite set $A$ of integers with $|A|\ge 2$, we have
$$
\int|R|^4\,d\mu\le \gamma\_2|A|^3
$$
with some absolute $\gamma\_2<1$ where $R(z)=\sum\_{a\in A}z^a$


This last part is easier to do combinatorially. Let $N=|A|\ge 2$. For every $k\in\mathbb Z$ define $D\_k=\#\{a\in A:a+k\in A\}$. Then the LHS is just $\sum\_k D\_k^2$. Notice that $\sum\_k D\_k=N^2$ and $D\_k\le N$. Thus, the only way not to have a constant improvement over the trivial $N^3$ bound is to have at least $0.9N$ integers $k$ for which $D\_k>0.9N$. Let $B$ be the set of such $k$. Then $|B|\ge 0.9N$. However if $k,\ell\in B$, then
$$
D\_{k-\ell}\ge \#\{a\in A:a+k\in A, a+\ell\in A\}\ge 0.8N 
$$
Thus $B-B\subset C$ where $C=\{k:D\_k\ge 0.8N\}$. However, by the identity $\sum\_k D\_k=N^2$, we get $|C|\le 1.25N<1.8N-1\le 2|B|-1$, which is impossible
(this is a trivial sumset theorem: if $B=\{b\_1<b\_2<\dots<b\_L\}$, then
$B-B\supset\{b\_1-b\_L<b\_1-b\_{L-1}<b\_2-b\_{L-1}<\dots<b\_L-b\_1\}$).


