Almost there?? Help with inequality.(updated) I was recently doing a combinatorics problem from MMO given in book Arthur Engel. I might have solved it (though unlikely) as I arrived at following inequality:-
Both $x$ and $y$ are positive. the first term is not in the pattern, while rest are.
$($$\frac{(x^3+y^3)}{(x+y)^3}$$)$ + $($$\frac{(x^4+y^4+x^3y+y^3x+x^2y^2)}{(x+y)^4}$ + $\frac{(x^5+y^5+x^4y+y^4x+x^3y^2+x^2y^3)}{(x+y)^5}$ ....$)$ $<1$ ?
I don't know if it is true (not good with inequalities). Please verify it.
Any help is appreciated :)
I can post the question if you want to give it a try but please don't tell the solution here.

My method:
$(1)$ All triangles are similar and each triangle has unique a unique area. Two triangles are congruent iff their areas are equal. Then, I constructed an ''area-based representation''
$(2)$ If the bigger RAT(Right angled triangle) into area $x$ and $y$, Then when we divide the RAT with area $x$ we end up with triangles of areas: $x^2/(x+y)$ and $xy/(x+y)$. So if we just keep dividing the RAT (not caring about congruency at this point)  We end up with the nth row of pascals triangle/$(x+y)^{(n-2)}$.
                  1(x+y)
               1(x)  1(y)
         1(x^2/(x+y)) 2(xy/(x+y)) 1(y^2/(x+y))
 1(x^3/(x+y)^2) 3(x^2y/(x+y)^2)  3(xy^2/(x+y)^2) 1(y^3/(x+y)^2)

$(3)$ If we keep the first triangle unchanged, then divide the second triangle into $x$ and $y$, then divide the third into $x^2/(x+y)$,$y^2/(x+y)$,$xy/(x+y)$,$x^2y/(x+y)^2$,$xy^2/(x+y)^2$, then the fourth triangle must be
$($$\frac{(x^3+y^3)}{(x+y)^2}$$)$ + $($$\frac{(x^4+y^4+x^3y+y^3x+x^2y^2)}{(x+y)^3}$ + $\frac{(x^5+y^5+x^4y+y^4x+x^3y^2+x^2y^3)}{(x+y)^4}$ ....$)$
which I wanted to show is less than $x+y$
$Attempt(promising?)$
I noticed that in case of maximum area we can't get $(x^4,y^4,x^5,y^5...)$ , then I mordified the inequality but was only able to give the required proof for $0.276 < x <  0.723 $
Reason:  Notice that when we are dividing one single triangle we can get many terms but from the each of the sets $S_x$ and $S_y$ we can only get one term from each ( so in total $2$ distinct terms and think about this yourself )
$S_x = (1,x,x^2,x^3,x^4...) $
$S_y = (1,y,y^2,y^3,y^4...) $
And since there are four triangles we can get (in max area case) ${1,x,y,x^2,y^2,x^3,y^3}$
 A: Assuming your LHS is correct, you can assume from homogeneity and symmetry, WLOG let $t= x/y > 1$:
$$\frac{x^3+y^3}{(x+y)^3}+\sum_{n=5}^\infty \frac{\sum_{k=0}^n x^{n-k}y^k}{(x+y)^n}  = \frac{t^3+1}{(t+1)^3}+ \sum_{n=5}^\infty \frac{\sum_{k=0}^nt^k}{(t+1)^n} = f(t)$$
Then (using GP formula, you can / should do the working yourself)
$$f(t)=\frac{t^2-t+1}{(t+1)^2}+\frac1{t-1}\sum_{n=5}^\infty\left( t\cdot\left(\frac{t}{t+1}\right)^n-\frac1{(t+1)^n}\right) \\= \frac{t^6+2t^5+2t^4+t^3+2t^2+2t+1}{t(t+1)^4}$$
It is clear that $f(t)$ increases without bound (as numerator is a higher degree polynomial), and in fact even$f(2)= \dfrac{181}{162}>1$.
A: (This is more of a comment regarding OP's approach to the original question. It's not a complete solution.)
With regards to your method, you're on the right track.

*

*Note that $ x + y = 1$, so you could get rid of that.

*Note that the triangles are split up into areas of size $ x^a y^b$.

*For $ (x, y) \neq (1/2, 1/2)$, 2 triangles are congruent iff $ a_1 = a_2, b_1 = b_2$.

*

*Hence, if we have distinct triangles, then their areas will be a subset of $ 1, x, y, x^2, xy, y^2, x^3, x^2y , xy^2, y^3, ... $

*Notice that this  is different from your value.



*For $ (x, y) = (1/2, 1/2)$, w triangles are congruent iff $a_1 + b_1 = a_2 + b_2$.

*

*(For now, I will ignore this case and leave it to you).



