Prove that if \begin{eqnarray} a_1&\leq& b_1\leq c_1,\\ a_2&\leq& b_2\leq c_2,\\ &\vdots&\\ a_N&\leq& b_N\leq c_N, \end{eqnarray} then \[ \sum_{i=1}^Na_i\leq \sum_{i=1}^Nb_i\leq \sum_{i=1}^Nc_i. \]

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Lesson Parent: 
Rules of Inequalities
Problem: 

Prove that if
\begin{eqnarray}
a_1&\leq& b_1\leq c_1,\\
a_2&\leq& b_2\leq c_2,\\
&\vdots&\\
a_N&\leq& b_N\leq c_N,
\end{eqnarray}
then
\[
\sum_{i=1}^Na_i\leq \sum_{i=1}^Nb_i\leq \sum_{i=1}^Nc_i.
\]

Answer: 

It is true that if \begin{eqnarray} a_1&\leq& b_1\leq c_1,\\ a_2&\leq& b_2\leq c_2,\\ &\vdots&\\ a_N&\leq& b_N\leq c_N, \end{eqnarray} then \[ \sum_{i=1}^Na_i\leq \sum_{i=1}^Nb_i\leq \sum_{i=1}^Nc_i. \]

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.