Oppa Putnam style

I'm French, 26, currently working as a structural engineer in Geneva.
As far as I can remember math has always been what I love to do. In my teens I would spend hours and days and months of my spare time on problems trying to invent new math theorems. I was naive enough to think that would be possible at my level but it was still fun and I did "invent" a couple of things (that had already been invented but still ...).
I kept on doing as much math as I could after high school, which in France means "classes préparatoires" and then "Grandes Ecoles", and successfully passed the exam for one of the hardest school in France (hardest to get in to that is). We always have shitty rating in international rankings because we are a "small" school (500 students per year) but lately we ranked 4th world in the Times Higher Education Alma Mater Index (http://www.timeshighereducation.co.uk/news/alma-mater-index-global-executives-2013/2007032.article).
At that time (20 years old) I still had no clue what I wanted to do with my life. I knew I didn't want to be a teacher, nor a researcher, and I hate finance in general so I saw little hope for me of doing something I like related to pure mathematics. A shame as I was pretty decently good in math and not so good in physics, mechanics etc... So I sold out and learnt some "practical" skills, and here I am today, working as a civil engineer. It's a great deal of fun mind you, but maths will always be in my heart.

The answer is obviously n (testing on a couple numbers)
I first proved that each solution for a given n that had the same number k of integers was unique. First terms are equal otherwise one is stricly bigger than the other, etc...
Then I found a way to write n as the sum of k integers for each k between 1 and n with an euclidean division. n=pk+r=(k-r)*p + (r)*(p+1).

My answer kinda sucks for this one, I start by simplifying the problem by dividing everything by the a_i so I have a problem only in (1+c_i) etc. (where c_i is b_i/a_i). Then I just developed everything and found that solving the problem was equivalent to proving that an nth geometrical average of n terms is always smaller than an arithmetic average of the same n terms. Which is a known fact. I had forgotten how to prove this so I proved it by replacing one of the terms with x and deriving. The problem is then solved by mathematical induction. (I think you can also prove that with concave functions).

I spent A LOT of time on this one. I'm still convinced that there is an easy solution I didn't see... So in desperation, I wrote out A(x) = cos(x) + 1/cos(x) + sin(x) + 1/sin(x) + tan(x)+ 1/tan(x) as a function of t=tan(x/2). After developing I find A(x)=B(t)=-2+1/t+(t-1)/(t^4-1). I then try to solve B'(t) = 0 which leads me to t^4-4t^3-1 = 0. I use Ferrari's method (first time this has ever been useful to me) to factor. This leads me to solving the equation (I forget what the solution is exactly). Then by cleverly rewriting B(t) with t^4-1 = -4t^3 and using the second degree equation I found to write 1/t as at+b, I find that the answer is 1-2*sqrt(2).

This on was surprinsingly easy.
I started by proving that if abs(P(x)) <= abs(Q(x)) than
Q(x)=0 had no real answers (unless they were the same as P(x) in which case the problem is obvious.
Then I solved the problem if Q(x)=0 has no real answers. Easily enough, the minimum of P is smaller than the minimum of Q and the minimum of a second degree polynom is Delta/(2a). aP<aQ
(if we consider both polynoms to be positive (convention) and the answer is obvious.)
You just have to realize then that solving the problem is equivalent to proving that P(x)-Q(x)=0 has no answer and P(x)+Q(x)=0 has no answers. Which is easily done by looking at the Delta for P-Q and P+Q which is always negative (one or the other).

The convention I use is that a path is written as follows (1,-1,1,1,1,-1,-1,-1) for example.
All the Dyck n-path can be written as a unique series of smaller Dyck k-path.
is s is a Dyck n-path, s=a1,a2,...,ap with ai irreductible Dyck paths (irreducticle meaning with only one return).
The function F(s) I used to link Dyck n-1 path ---> Dyck n patch with only odd returns is
For all s in Dyck(n-1), s=a0,a1,...,ak-1,b,ak+1,...,ap where b is the first (starting from the right) Dyck path with an even return. b can be () the null Dyck path. Then F(s)=(1,a0,...,ak-1,-1,ak+1,...,ap. I then had to prove that this was indeed a bijection (which was a bit tedious) but relatively easy.

For this one, I started proving that if it existed, it was unique. I proved this by mathematical induction by building the partitions from 0.
1 and 0 can't be in the same partition otherwise 1=a+b would have 1 solution in that partition and zero in the other. By building up, at step n you realize that you need to place n in one or the other partition based on how many solution a+b=n already has in each partition.

Then I proved that the group existed by noticing that the solution pairs for n=x+y used all the numbers from 0 to n-1 (one and only once) if n is odd, and all the numbers from 0 to n-1 except n/2 if n is even. You can then notice that for each n, whatever the partitions of equal sizes, of (1,2,...,n-1) if n is even, then if you put n in the partition with n/2 they will have the same number of solutions. If n is odd, the partitions will have the same number of solutions if you place n in the partitions in the smallest partition.