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. xi<xj x_i < x_j and yi<yj,i<j y_i < y_j, \forall i < j ,

  2. 1<x1<x2<...<xn<y1<...<ym, 1<x_1<x_2<...<x_n<y_1<...<y_m,

  3. x1+x2++xn>y1++ym.x_1+x_2+ \ldots +x_n > y_1+ \ldots +y_m.

Prove that: x1×x2××xn>y1×y2×ym x_1 \times x_2 \times \cdots \times x_n > y_1 \times y_2 \times \cdots y_m.

#Algebra #Smoothing

Note by Calvin Lin
7 years ago

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Comments

can we proove this by taking log in the proove part and as log is an increasing function the following can be true ?

Ashish Nagpal - 7 years ago

Here's my approach.

Prove the following version instead.

For two positive integer x1,x2,...,xnx_1,x_2,...,x_n and y1,y2,...,ymy_1,y_2,...,y_m which satisfy

  1. xi<xj x_i < x_j and yi<yj,1<i<j y_i < y_j, \forall 1 < i < j ,

  2. 1<x2<...<xn<y2<...<ym, 1<x_2<...<x_n<y_2<...<y_m, and 1x1 1 \leq x_1 and 1y1 1 \leq y_1

  3. x1+x2++xn>y1++ym.x_1+x_2+ \ldots +x_n > y_1+ \ldots +y_m.

Prove that: x1×x2××xny1×y2×ym 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 x1,y1 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=1m=1 separately. In particular, it leads to the only equality cases. Proving m=1m=1 in the original question is straightforward.

Calvin Lin Staff - 7 years ago

You can use Abel summation techniques. :)

A Brilliant Member - 7 years ago

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Interesting. Can you elaborate?

Calvin Lin Staff - 7 years ago

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

Ramesh Goenka - 7 years ago

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

Calvin Lin Staff - 7 years ago
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