Solution of problem 2, International Mathematical Olympiad 2020
1- What is IMO first?
The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads, each year is organized in a country and several hundred of students at the end of high school participate. The selection process differs by country, but it often consists of a series of tests which admit fewer students at each progressing test. Awards are given to approximately the top-scoring 50% of the individual contestants.
The content ranges from extremely difficult algebra and pre-calculus problems to problems on branches of mathematics not conventionally covered at school and often not at university level either, such as projective and complex geometry, functional equations, combinatorics, and well-grounded number theory.
source wikipedia
I like spending some of my free time solving some IMO problems because I find them very fun, given their nature which prioritizes simple and intuitive thinking.
2- Problem statement
3- Solution and discussion
the 4 real numbers composing this inequality are already ordered, and their sum is equal to 1 (we need to use these pieces of information well), so looking at the inequality quickly, the part that is a little unusual or a little difficult to parse is where the numbers are raised to the power of themselves. Thus, the first step of my solution was to simplify that part.
To do this, we need to use the famous weighted version of “Inequality of arithmetic and geometric means” which says that :
Theorem :
let the non negative numbers x1, x2, . . . , xn and the non negative weights w1, w2, . . . , wn be given. Set w = w1 + w2 + · · · + wn. If w > 0, then the inequality:
in our case we have exactly the same thing, with n=4 and
w1 = x1= a, w2 =x2= b, w3 =x4= c and w4 =x4= d
by directly applying the inequality we have that :
since w= w1+w2+w3+w4 = a+b+c+d=1 so we have:
using the result of inequality (1), now we have just to demonstrate that :
At this stage the observation that we must have here is that if we expand the left-hand side of the inequality, we will have terms in powers of 3 with other product terms. It looks like a third degree expansion of (a+b+c+d)³, guess what (a+b+c+d)³=1 !.
what you need to know about is that:
I know that the formula of third degree binomial expansion with 4 terms is not very common, but it can be obtained with a simple calculation with binomial expansion of 2 terms.
At this point what we need to do, is playing with terms using the information that a≥b≥c≥d>0 in some way to get part of the right-hand side of equality (2).
Our play strategy is to try to isolate the numbers a, b, c and d at each step :
a+2b+3c+4d ≤a+3b+3c+3d (3) => because b≥d
a+2b+3c+4d≤3a+b+3c+3d (4) =>because a≥b
a+2b+3c+4d≤3a+3b+c+3d (5) =>because b≥c
a+2b+3c+4d≤3a+3b+3c+d (6) =>because c≥d
So if we do the following computation
a²*(3)+b²*(4)+c²*(5)+d²*(6) => we will have the following inequality:
with x1=a, x2=b, x3=c and x4=d
using equation (2) we know that :
Finally we can conclude from (7) and (8) that
(a+2b+3c+4d)(a²+b²+c²+d²)<(a+b+c+d)³=1
using the result (1) we have the final inequality
Thank you for reading! and please let me know if you find any other alternative solutions.
