A sum over a hyperplane in $\mathbb{Z}^4$ 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: 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
$B-B\supset\{b_1-b_L<b_1-b_{L-1}<b_2-b_{L-1}<\dots<b_L-b_1\}$).
