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I'm really bad when it comes to proving inequalities. I have prove this:

these are all positive

$\sum\limits_{k=1}^n a_k \leq \sum\limits_{k=1}^n ka_k \leq n \sum\limits_{k=1}^n a_k$

Where would i start with this? Anyone have a simple example i they can show me step by step. I just don't understand how to prove inequalities.

thanks!


Edit

So, would this work: $\sum_{k=1}^\left(n+1\right) a_k = \sum_{k=1}^n a_k + a_\left(k+1\right)$

then subtracting you would get

$2a_\left(k+1\right)$

Now for the right side, do the same thing to get

$2ka_\left(k+1\right)$

Which i now have

$2a_\left(k+1\right) \leq 2ka_\left(k+1\right)$

which reduces to

$1 \leq k$

That finishes the proof for that portion. Am i on the right track for this?

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    @Lovre Im going to update my post right now, im not sure if i did this correctly, but it's my attempt at it.2011-04-22

3 Answers 3

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Start with $1\le k \le n$. Just to be sure, are you assuming $a_k \ge 0$, for all $k$? Are the $a_k$ the antecedent ''these'' in your statement?

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    it just says prove that if a1, a2, a3,...,an are positive then (The summation inequalities), but im going to update my post to see if im getting this right.2011-04-22
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I'm assuming you mean that $a_k \geq 0$ for all $k$. Here's a hint.

Try to approach the inequalities term by term i.e.,

\begin{equation} a_k \leq b_k \leq c_k\end{equation}

Then the inequalities will be maintained under summation, which should give you your result.

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    I had trouble understanding your edit. If you want to prove it by induction, look at Ross Millikan's answer. I personally think induction in this problem is a long drawn out way of doing what can be accomplished in two or three lines.2011-04-22
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One approach is by induction, which it looks like you are trying. Under Edit you say "then subtracting" but don't say what you are subtracting. If you are going to prove it by induction, you need to prove a) it is true for $n=1$; and b) if it is true for a given $n$, then it is true for $n+1$. Working on the left hand inequality, for $n=1$ it says $\sum_{k=1}^1 a_k \leq \sum_{k=1}^1 ka_k$, which is true because $a_1\leq 1a_1$. Now if we know $\sum_{k=1}^n a_k \leq \sum_{k=1}^n ka_k$, $\sum_{k=1}^{n+1} a_k=a_{n+1}+\sum_{k=1}^n a_k \leq(n+1)a_{n+1}+ \sum_{k=1}^n ka_k \leq =\sum_{k=1}^{n+1} ka_k$, so it is true for $n+1$. You should be able to do the other inequality similarly.

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    Because on the right hand side you had $n$ multiplying the sum when you added $n$ terms and that part of the sum is now multiplied by $n+1$ when you sum $n+1$ terms.2011-04-22