I've come across the following question on the internet:
For two positive integer \(x_1,x_2,...,x_n\) and \(y_1,y_2,...,y_m\) which satisfy
xi<xj and yi<yj,∀i<j,
1<x1<x2<...<xn<y1<...<ym,
x1+x2+…+xn>y1+…+ym.
Prove that: x1×x2×⋯×xn>y1×y2×⋯ym.
#Algebra
#Smoothing
Easy Math Editor
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can we proove this by taking log in the proove part and as log is an increasing function the following can be true ?
Here's my approach.
Prove the following version instead.
For two positive integer x1,x2,...,xn and y1,y2,...,ym which satisfy
xi<xj and yi<yj,∀1<i<j,
1<x2<...<xn<y2<...<ym, and 1≤x1 and 1≤y1
x1+x2+…+xn>y1+…+ym.
Prove that: x1×x2×⋯×xn≥y1×y2×⋯ym.
This version is much easier to work with. We then prove strict inequality by looking at the equality cases.
Hint: Think about what x1,y1 could be made to do. Why would I want to make them special?
Hint: How would you minimize the LHS and maximize the RHS? Try smoothing.
Hint: Deal with m=1 separately. In particular, it leads to the only equality cases. Proving m=1 in the original question is straightforward.
You can use Abel summation techniques. :)
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Interesting. Can you elaborate?
i think the condition n>m must be mentioned if all are positive integers greater than 1 .. !
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That can be deduced from the conditions. In fact, I believe we have m<n<2m.