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). 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 (although there has been some other progress since then).
Szegedy and Li gives a different proof of the above inequality, using convexity of the logarithm function.
Also see the paper of Kim, Lee, and Lee for another approach.
On page 28 of Lovasz' book on graph limits, 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
Best Answer
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*}