Prove that if a1≤b1≤c1,a2≤b2≤c2,⋮aN≤bN≤cN, then N∑i=1ai≤N∑i=1bi≤N∑i=1ci.

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Lesson Parent: 
Rules of Inequalities
Problem: 

Prove that if
a1≤b1≤c1,a2≤b2≤c2,⋮aN≤bN≤cN,
then
N∑i=1ai≤N∑i=1bi≤N∑i=1ci.

Answer: 

It is true that if a1≤b1≤c1,a2≤b2≤c2,⋮aN≤bN≤cN, then N∑i=1ai≤N∑i=1bi≤N∑i=1ci.

SAMPLE SOLUTION

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Solution: 

For many students this is one of those statements that is very obvious. We agree that it is. However, to formally prove it takes a few steps. The proof will be by induction where we prove that it is true for N=2, assume it is true for some arbitrary n and then prove it is true for n+1.

We have shown here that this is true for N=2.

Now let's assume this is true for some arbitrary n. In other words, we will assume that since

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is true that

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is also true.

Now let's prove that if

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is true, then

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is also true. Let's first define the following symbols:

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So we know that <Sign in to see all the formulas>. But because we also know that <Sign in to see all the formulas> we can use this problem to show that

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Therefore,

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is true and the proof is complete.