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

\( x_i < x_j\) and \( y_i < y_j, \forall i < j \),

\( 1<x_1<x_2<...<x_n<y_1<...<y_m, \)

\(x_1+x_2+ \ldots +x_n > y_1+ \ldots +y_m.\)

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

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TopNewestcan we proove this by taking log in the proove part and as log is an increasing function the following can be true ? – Ashish Nagpal · 3 years, 1 month ago

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You can use Abel summation techniques. :) – Paramjit Singh · 3 years, 1 month ago

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– Calvin Lin Staff · 3 years, 1 month ago

Interesting. Can you elaborate?Log in to reply

Here's my approach.

Prove the following version instead.

For two positive integer \(x_1,x_2,...,x_n\) and \(y_1,y_2,...,y_m\) which satisfy

\( x_i < x_j\) and \( y_i < y_j, \forall 1 < i < j \),

\( 1<x_2<...<x_n<y_2<...<y_m, \) and \( 1 \leq x_1 \) and \( 1 \leq y_1 \)

\(x_1+x_2+ \ldots +x_n > y_1+ \ldots +y_m.\)

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. – Calvin Lin Staff · 3 years, 1 month ago

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i think the condition n>m must be mentioned if all are positive integers greater than 1 .. ! – Ramesh Goenka · 3 years, 1 month ago

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– Calvin Lin Staff · 3 years, 1 month ago

That can be deduced from the conditions. In fact, I believe we have \( m < n < 2m \).Log in to reply