Is there a "Cauchy-Schwarz proof" of the following inequality?


**Theorem.** Given $f \colon [0,1]^2 \to [0,1]$, one has
$$
\int\_{[0,1]^4} f(x,y)f(z,y)f(z,w) \, dxdydzdw \geq \left(\int\_{[0,1]^2} f(x,y) \, dxdy\right)^3.
$$


*Background.*
This inequality is due to [Blakley and Roy (1965)](http://www.ams.org/mathscinet-getitem?mr=184950). In fact, they proved even more, namely when the LHS corresponds to a path of length $k$ (above $k=3$) and the RHS is $(\int f)^k$. 


This is a special case of a more general Sidorenko's conjecture, which claims that $t(H,W) \geq (\int W)^{e(H)}$ for any bipartite graph $H$. The general case of Sidorenko's conjecture is still open. See, e.g., this [note by Conlon, Fox, and Sudakov](http://math.mit.edu/~fox/paper-sidorenko-short.pdf) (although there has been some other progress since then).


[Szegedy and Li](http://arxiv.org/abs/1107.1153v1) gives a different proof of the above inequality, using convexity of the logarithm function. 


Also see the paper of [Kim, Lee, and Lee](http://arxiv.org/abs/1310.4383) for another approach.


On page 28 of [Lovasz' book on graph limits](https://rads.stackoverflow.com/amzn/click/com/0821890859), it states this inequality without proof, and then says



> 
> ... and this is already quite hard, although short proofs with a tricky application of the Cauchy–Schwarz inequality are known.
> 
> 
> 


So my question is: how does one prove the inequality above using Cauchy-Schwarz?


*Update:* It has been shown that there is no vanilla sum-of-squares proof of the inequality <https://arxiv.org/abs/1812.08820>



Cauchy-Schwarz proof of Sidorenko for 3-edge path (Blakley-Roy inequality)
Before giving a very short Cauchy-Schwarz inequality proof for the 3-edge path (can be done in a similar fashion for any tree), let me comment on the authorship of the inequality in question.


In 1959, Mulholland and Smith <https://doi.org/10.2307/2309342> proved that for any symmetric non-negative matrix $A$ and any non-negative vector $z$ of the same order
$$
 (z^{\*} A^k z) \cdot 
 (z^{\*} z)^{k-1}
 \; \geq \;
 (z^{\*} A z)^k \; ,
$$
where equality takes place if and only if $z$ is an eigenvector of $A$ or a zero vector.


Almost at the same time, Atkinson, Watterson and Moran <https://doi.org/10.1093/qmath/11.1.137> proved
$
 nm \cdot s(A A^{\*} A) \; \geq \; s(A)^3
$
for an asymmetric non-negative $(n \times m)$-matrix $A$,
and conjectured that a similar inequality holds for a $k$-edge path with $k>3$.
(Here $s(A)$ denotes the sum of entries of $A$).
They presented this inequality in both matrix and integral form.


Being unaware of these results, Blakley and Roy in 1965 published theirs: <https://doi.org/10.1090/S0002-9939-1965-0184950-9>.


Here is my favorite proof of the "Mulholland-Smith / Atkinson-Watterson-Moran / Blakley-Roy inequality".
Let $g(x) = \int f(x,y) \: dy$.
Then
\begin{multline\*}
 \int\int\int\int f(x\_1,y\_1) \; f(x\_2,y\_1) \; f(x\_2,y\_2) 
 \; dx\_1 \; dx\_2 \; dy\_1 \; dy\_2
 \\
\shoveleft
 = \int\int\int f(x\_1,y\_1) \; f(x\_2,y\_1) \; g(x\_2) \; dx\_1 \; dx\_2 \; dy\_1
 \\
\shoveleft
 \geq \int\int\int g^{1/2}(x\_1) \; f(x\_1,y\_1) \; f(x\_2,y\_1) \; g^{1/2}(x\_2)
 \; dx\_1 \; dx\_2 \; dy\_1
 \\
\shoveleft
 = \int \left( \int g^{1/2}(x) \; f(x,y) \; dx \right)^2 \; dy
 \\
\shoveleft
 \geq \left( \int\int g^{1/2}(x) \; f(x,y) \; dx \; dy \right)^2
 \; = \; \left( \int g^{1/2}(x) \; g(x) \; dx \right)^2
 \\
\shoveleft
 \geq \left( \int g(x) \; dx \right)^3
 \; = \; \left( \int f(x,y) \; dx \; dy \right)^3 \; .
 \\
\end{multline\*}


