Let ABCD be a tetrahedron of volume 1 and M,N,P,Q,R,S on AB,BC,CD,DA,AC,BD s.t. MP,NQ,RS are concurrent. Then the volume of MNRSPQ is less than 1/2. So far, I have denoted
$$\frac{MA}{MB}=a,\frac{BN}{NC}=b,\frac{CP}{PD}=c,\frac{DQ}{QA}=d,\frac{AR}{RC}=e,\frac{BS}{SD}=f$$
and then I have computed the volumes of the "small" tetrahedra
$$V_1=V_{AMRQ}=\frac{ae}{(a+1)(e+1)(d+1)};$$
$$V_2=V_{BMSN}=\frac{bf}{(b+1)(f+1)(a+1)};$$
$$V_3=V_{CRNP}=\frac{c}{(c+1)(e+1)(b+1)};$$
$$V_4=V_{DPQS}=\frac{d}{(d+1)(c+1)(f+1)}.$$
Also, by Menelaus theorem for tetrahedra, one can deduce that
$$abcd=1;~~~ af=ce; ~~~b=fde.$$
It remains to show that $$V_1+V_2+V_3+V_4\geq \frac{1}{2}.$$
The problem has a well known analogue in 2-D: for a triangle $ABC$ of area 1 and $AM, BN, CP$ concurrent cevians, the area of the triangle $MNP$ is less than 1/4.

Later edit/update:
I have managed to rewrite $V_k$ as follows:
$$V_1={\frac {a}{ \left( a+1 \right)  \left( d+1 \right)  \left( cd+1
 \right) }};$$
$$V_2={\frac {b}{ \left( a+1 \right)  \left( b+1 \right)  \left( ad+1
 \right) }};$$
$$V_3= {\frac {c}{ \left( c+1 \right)  \left( b+1 \right)  \left( ab+1
 \right) }};$$
$$V_4= {\frac {d}{ \left( c+1 \right)  \left( d+1 \right)  \left( bc+1
 \right) }},$$
Plus, we still have $abcd=1$.
 A: Let $X$ be the intersection of the three segments $MP$, $NQ$, $RS$.
Since $ABCD$ is non-degenerate, its volume is one, the points $A,B,C,D$ are not in a plane, so we can uniquely write
$$
X = aA+bB+cC+dD\ , \qquad a+b+c+d=1\ .
$$

Then grouping pairs of vertices, we obtain formulas for $M,N,P,Q,R,S$ in terms of these weights, corresponding to the groupings:
$$
\begin{aligned}
X 
&= 
(a+b)\cdot \underbrace{\frac {aA+bB}{a+b}}_{M} + 
(c+d)\cdot \underbrace{\frac {cC+dD}{c+d}}_{P} 
\\[2mm]
&= 
(a+c)\cdot \underbrace{\frac {aA+cC}{a+c}}_{R} + 
(b+d)\cdot \underbrace{\frac {bB+dD}{b+d}}_{S} 
\\[2mm]
&= 
(a+d)\cdot \underbrace{\frac {aA+dD}{a+d}}_{Q} + 
(b+c)\cdot \underbrace{\frac {bB+cC}{b+c}}_{N} \ .
\end{aligned}
$$
(They really know in Romania vectors in 3D in the VIII.th class. The relations
the children could have been written involve the vectors $OX$, $OA$, $OB$, $OCV$, $OD$ instead of the above.)
So $M$ is situated on $AB$ such that $M=\frac 1{a+b}(aA+bB)$, which gives
$AM:AB=b:(a+b)$. We write similar relations for $AR:AC=c:(a+c)$ and $AQ:AD=d:(a+d)$. So the volume $[AMRQ]$, seen as a fraction of the volume $[ABCD]=1$
is
$$
[AMRQ]
=
\frac{AM}{AB}\cdot
\frac{AR}{AC}\cdot
\frac{AQ}{AD}\cdot
[ABCD]
=\frac{bcd}{(a+b)(a+c)(a+d)}\cdot 1\ .
$$
So the given geometric inequality is equivalent to the algebraic inequality
$$
\sum\frac{bcd}{(a+b)(a+c)(a+d)}
\ge
\frac 12\ .
$$
The sum has four terms obtained by acting with the powers of the cyclic permutation  $(a,b,c,d)$ on the written term.

I could find quickly only the brute force proof. We multiply with the common denominator and have to show the domination:
$$
\underbrace{\sum a^3b^2c}_{4\cdot 3\cdot 2=24\text{ terms}} +
2\underbrace{\sum a^2b^2c^2}_{4\text{ terms}} 
\ge 
2\underbrace{\sum a^3bcd}_{4\text{ terms}} +
4\underbrace{\sum a^2b^2cd}_{6\text{ terms}} 
\ .
$$
The domination is clear, the monominal type $(3,2,1,0)$
dominates $(3,1,1,1)$ and $(2,2,1,1)$. And $(2,2,2,0)$ dominates $(2,2,1,1)$.
So one can stop here.
But for the convenience of the reader, let us write this explicitly. First of all
$$
b^2c+c^2b+b^2d+d^2b + c^2d+d^2c\ge 6bcd\ .
$$
(Arithmetic-geometric mean.) We multiply with $a^3$ and sum. It remains (i.e. is sufficient) to show
$$
\frac 23
\underbrace{\sum a^3b^2c}_{4\cdot 3\cdot 2=24\text{ terms}} +
2\underbrace{\sum a^2b^2c^2}_{4\text{ terms}} 
\ge 
4\underbrace{\sum a^2b^2cd}_{6\text{ terms}} 
\ .
$$
Now use $a^3b^2c+ab^2c^3\ge 2a^2b^2c^2$ to estimate downwards the LHS with a scalar multiple of $\sum a^2b^2c^2$, to have a simpler LHS, so
$$
\begin{aligned}
\frac 23
\underbrace{\sum a^3b^2c}_{4\cdot 3\cdot 2=24\text{ terms}} +
2\underbrace{\sum a^2b^2c^2}_{4\text{ terms}} 
&\ge 
\frac 23\cdot 6
\underbrace{\sum a^2b^2c^2}_{4\text{ terms}} +
2\underbrace{\sum a^2b^2c^2}_{4\text{ terms}} 
=
6
\underbrace{\sum a^2b^2c^2}_{4\text{ terms}} 
\\
&\ge
4\underbrace{\sum a^2b^2cd}_{6\text{ terms}} 
\ .
\end{aligned}
$$
The last inequality follows from $a^2b^2c^2+a^2b^2d^2\ge 2a^2b^2cd$.
$\square$

Note: The problem was proposed by Flavian Georgescu, Bucharest, at that time college student (elev), so maximally 18yo, and it should be doable with the aid of "thin air" available in the 8.th class (14yo students) of the Romanian mathematical matter from the school book. I suppose Titu or a mean inequality was the solution of the author, if he did it as above, else i see only the possibility of a geometrical argument using $[XMRQ]$ and the similar tetraedra.
A: Well this not a rigorous proof for this problem, but I decided to express some of my ideas that may lead to a geometric proof.
First I managed to find set of points of concurrency $S_i$ inside an arbitrarily triangle $T$ which make same area of triangles shaped by their three concurrent segments as you mentioned in your 2-D case. the sets of such points can also be defined by the level sets of a function $f$ that maps each point inside of the triangle $T$ to the area of its mentioned triangle. by use of continuity we can deduce that every level set is a jordan closed loop $J_i,i\in [0,1]$ and even it seems that they are convex loops which doesn't intersect eachother and their value by function $f$ (which resembles the mentioned triangles area) become bigger and bigger as the loops become smaller and smaller until they retract to centroid of triangle $T$.
I think this process can be easily generalized to the case of the the 3-d tetrahedron and the level sets here would become border of convex bodies which retracts to the centroid of tetrahedron when it gains its highest value by the mapping function sending each point inside of tetrahedron to the volume of its octahedron you mentioned.
While searching for a proof of this claim ,I was curious about a transformation that convert an arbitrarily tetrahedron to a regular one as it seems good idea too for the 2-D case to convert a given triangle to the most symmetrical triangle which is equilateral triangle. I've just find a kind of such transformation here. I still don't know this transformation would be useful from the perspective of my approach for the sake of the statement of inequality of areas in this problem.
for example if the jordan loops $J_i$ in $T$ becomes circles while transformed, then the proof would become easier.
My second idea is about to prove this statement:

The volume of mentioned octahedron $MNPQRS$ will be maximum if and only if the centroids
of $ABCD$ and $MNPQRS$ coincide.

