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# Smoothing An Inequality

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

1. $$x_i < x_j$$ and $$y_i < y_j, \forall i < j$$,

2. $$1<x_1<x_2<...<x_n<y_1<...<y_m,$$

3. $$x_1+x_2+ \ldots +x_n > y_1+ \ldots +y_m.$$

Prove that: $$x_1 \times x_2 \times \cdots \times x_n > y_1 \times y_2 \times \cdots y_m$$.

Note by Calvin Lin
3 years, 9 months ago

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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 ?

- 3 years, 9 months ago

You can use Abel summation techniques. :)

- 3 years, 9 months ago

Interesting. Can you elaborate?

Staff - 3 years, 9 months ago

Here's my approach.

For two positive integer $$x_1,x_2,...,x_n$$ and $$y_1,y_2,...,y_m$$ which satisfy

1. $$x_i < x_j$$ and $$y_i < y_j, \forall 1 < i < j$$,

2. $$1<x_2<...<x_n<y_2<...<y_m,$$ and $$1 \leq x_1$$ and $$1 \leq y_1$$

3. $$x_1+x_2+ \ldots +x_n > y_1+ \ldots +y_m.$$

Prove that: $$x_1 \times x_2 \times \cdots \times x_n \geq y_1 \times y_2 \times \cdots y_m$$.

This version is much easier to work with. We then prove strict inequality by looking at the equality cases.

Hint: Think about what $$x_1, y_1$$ 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.

Staff - 3 years, 9 months ago

i think the condition n>m must be mentioned if all are positive integers greater than 1 .. !

- 3 years, 9 months ago

That can be deduced from the conditions. In fact, I believe we have $$m < n < 2m$$.

Staff - 3 years, 9 months ago