# Proving inequalities

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

these are all positive

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

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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Is the problem mainly at getting the formal results, or do you have problem see why is the result the way it is? Do you see (informally) why the inequalities you posted are true? –  user5501 Apr 22 '11 at 3:14
@Lovre Im going to update my post right now, im not sure if i did this correctly, but it's my attempt at it. –  Matt Apr 22 '11 at 3:20

## 3 Answers

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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@Ross Millikan so how do you know its true. Thats what i don't understand about inequalities. I know you have to prove that 1 is true. Which i could see. Then you assume that n is true. So you have to prove for n+1. But from what you have above, how do you know the inequality is true? Thanks –  Matt Apr 22 '11 at 21:13
In my final line, I have a chain of $\le$ and $=$ relations from the LHS of the $\le$ to the RHS. It is like saying $2\le 3 = 3 \le 3 \le 4$ and concluding $2 \le 4$. Do you understand each step in the chain? Then I am using the transitivity of $\le$ –  Ross Millikan Apr 22 '11 at 21:17
@Ross Millikan - ok i think i see it now. So basically you are saying $\sum_{k=1}^{n+1} a_k \leq \sum_{k=1}^{n+1} ka_k$ and that would be the induction and proof portion. –  Matt Apr 22 '11 at 21:20
Exactly. I have only done the first of your two inequalities. The second is different, because when you evaluate the RHS for $n+1$ all the terms you had at $n$ increase, but you can do it. You would put in $\frac{n+1}{n}$ times the previous sum to $n$ as one of the steps in the chain. Really svenkatr's approach (which works on the first as well) is easier-just note that each term on the left side of the $\le$ sign is $\le$ the corresponding term on the right. –  Ross Millikan Apr 22 '11 at 21:29
@Ross Millikan - why would it be $\frac{n+1}{n}$ and not just $n+1$ ? –  Matt Apr 22 '11 at 21:51

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. –  Matt Apr 22 '11 at 3:21

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.,

$$a_k \leq b_k \leq c_k$$

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

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not sure what you mean. Could you look at the edit portion of my post and tell me if im doing it correctly? thanks –  Matt Apr 22 '11 at 4:35
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. –  svenkatr Apr 22 '11 at 5:29