Let $K$ be a centrally symmetric convex body in $\mathbb R^3$ with volume ${\rm vol}(K)=1$. For any subset $F \subset \lbrace1,2,3\rbrace$, let $K\_F$ be the projection of $K$ in $\mathbb R^F$.



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> **Question:** What is the best constant $C$, such that
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> $${\rm vol}(K\_{\lbrace 1 \rbrace}) \leq C \cdot {\rm vol}(K\_{\lbrace 1,2 \rbrace}) \cdot {\rm vol}(K\_{\lbrace 1,3 \rbrace}).$$
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> 
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Here, the volume of $K\_F$ is computed in $\mathbb R^F$. With some work one can prove that $C=2$ is good enough. This is elementary and follows for example from two applications of Lemma 3.1 in


J. Bourgain and V.D. Milman, New volume ratio properties for convex symmetric bodies in $\mathbb R^n$, Invent. Math. 1987 vol. 88 (2) pp. 319-340.


My guess is that maybe $C=1$ works, but I am not sure.



Volume inequality between projections of a convex symmetric set in $\mathbb R^3$
We can exchange the condition $\mathop{\rm vol}K=1$ to ${\rm vol}K\_{\lbrace 1 \rbrace}=1$.
In this case we need to show that 
$$\mathop{\rm vol}K\leqslant C\cdot \mathop{\rm vol}K\_{\lbrace 1,2 \rbrace} \cdot \mathop{\rm vol}K\_{\lbrace 1,3 \rbrace}.$$
The later is equivalent to the following:
$$\int\limits\_0^1 u{\cdot}v{\cdot}dx\leqslant C\cdot \int\limits\_0^1 u{\cdot}dx\cdot\int\limits\_0^1 v{\cdot}dx\ \ \ \ \ (\*)$$
for two positive convave functions $u,v\colon[0,1]\to\mathbb R$.
WLOG we can assume that both functions are PL and both bave $0$ at $0$ andat $1$.
Further we can assume that 
$$\int\limits\_0^1 u{\cdot}dx=\int\limits\_0^1 v{\cdot}dx=1. \ \ \ \ \ (\*\*)$$
We can move extremal points of $u$ and $v$ one by one keeping the identity $( \* \* )$ and increasing LHS of $( \* )$.
This way we may reduce number of extremal points in $u$ and $v$;
the process might terminate if you get one or two extremal point in each.


It seems that 
$$u=v=2-4\cdot|x-\tfrac 12|$$ 
is the worst case. So $C=\tfrac43$ is the best constant. 


