Trying to solve a system of inequalities I have this kind of system of inequalities (with one equal

starbright49ly

starbright49ly

Answered question

2022-05-31

Trying to solve a system of inequalities
I have this kind of system of inequalities (with one equality):
1. x + y + z = 1
2. a x + b y + c z 2
3. a 2 x + b 2 y + c 2 z 6
4. a 3 x + b 3 y + c 3 z 14
and so on.. (I could continue with 5. 6. 7.... the power always increases by one and I know the value on the right).
x,y,z are my unknowns and a,b,c are known.
Could I get a solution (or an approximation) for x,y and z as the number of inequalities grows, I don't see it?

Answer & Explanation

Danube2w

Danube2w

Beginner2022-06-01Added 7 answers

Not a full answer, but too long for a comment: let's first assume that all quantities involved are positive for convenience. Let u = ( x , y , z ), and let v n = ( a n , b n , c n ). The inequalities translate to the system
{ u 1 = 1 u v k b k , 1 k n
for given constants b k . By Cauchy-Schwarz, we have that u v k | u | | v k |, u v k | u | | v k | , so a sufficient (but certainly not necessary) condition for the system of inequalities to hold is | u | | v k | b k for each k, equivalently
x 2 + y 2 + z 2 b k 2 a 2 k + b 2 k + c 2 k .
Now, we run through 1 k n and calculate
m = min 1 k n { b k 2 a 2 k + b 2 k + c 2 k } .
Once we calculate this, then we are left with the restriction | u | < m which is an open ball in R 3 . On the other hand u 1 = 1 represents a plane, so taking any vector u which lies on their intersection is sufficient.
This is not a full solution, however: the method can fail because there might not be any u in their intersection, although there does exist u satisfying the original system. This is because of the use of Cauchy-Schwarz, which is not an if-and-only-if criterion.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?