# Smoothing Concept Example and Inequalities

1. Apr 23, 2007

### teleport

Hi, I am curious to kow about this smoothing concept that I've seen mentioned somewhere but do not understand. It would be great if someone could explain it to me, preferably through a solution to this problem. Thanks for any comments. (Latex did not generate in the preview, so if it seems there was a mistake with the code, please tell which was it.)

Let $$a_0,...,a_n$$ be real numbers in the interval $$(0, \frac{\pi}{2})$$ such that

$$$\tan (a_0 - \dfrac{\pi}{4})+...+\tan (a_n-\dfrac{\pi}{4})\geq n - 1$$$

Prove that

$$$\tan(a_0)\tan(a_1)...tan(a_n) \geq n^{n + 1}$$$

Last edited: Apr 23, 2007
2. Apr 23, 2007

### mathman

Unless there are further condintions on the a's, the statement is false. Let a0=0, the tan is then 0, while the first inequality can be made to hold by adjusting the other a's. The second expression will always have 0 for the left side.

3. Apr 23, 2007

### teleport

I am sorry, but the interval does not include 0, as it is open. If it did, it would also include $$\dfrac{\pi}{2}$$. Since tan at a = $$\dfrac{\pi}{2}$$ is undefined, it doesn't.

Last edited: Apr 23, 2007
4. Apr 30, 2007

### teleport

OK. If no one knows about the smoothing thing, then we might as well forget about it for the moment. Is there another type of solution? It's incredible how useful and easy to deal with, Jensen's inequality can be. But damn, we can't use it all the time!

5. May 2, 2007

### teleport

Well it does seem that mathematicians here are not that in touch with the name, or simply not interested. Anyways, I do find the concept,
now that I understand it much better, that in reality is simple, but highly useful. Nevertheless it may give the impression of being somewhat arbitrary for the one who sees its applications to solving problems, for the first time. For those who do not know much about it, or may have seen it but didn't understand like me (or no one could explain to you... ) here is a Lemma statement (probably a theorem, but I haven't seen it, yet, generalized for more than two numbers), and my
proof.

Lemma:

The product of any real numbers can be increased under the constraint i, if and only if the following condition ii, is met.

i) The sum of the numbers is fixed.

ii) The difference between the numbers is decreased.

Proof:

Let $$\left\{a_i\right\}_1^n$$ be a sequence of real numbers. Define the transform $$T:\left\{a_i\right\}_1^n\rightarrow \left\{\alpha\right\}_1^n$$ that will replace all the $$a_i$$ with $$\alpha = \displaystyle \sum_{i=1}^n \dfrac{a_i}{n}.$$
Under the transformation, $$\displaystyle \sum_{i=1}^n a_i$$ is fixed since

$$\displaystyle \sum_{i=1}^n a_i = \displaystyle \sum_{i=1}^n \alpha = n\alpha = \dfrac{n}{n}\displaystyle \sum_{i=1}^n a_i = \displaystyle \sum_{i=1}^n a_i$$
In addition, notice that the difference between any elements of $$\left\{\alpha\right\}_1^n$$ is zero or minimized, since all the $$\alpha_i$$ are equal. Hence, the conditions i,ii of Lemma are satisfied, and we only need to show

$$\displaystyle \prod_{i=1}^n \alpha > \displaystyle \prod_{i=1}^n a_i.$$

By AM-GM inequality

$$\begin{eqnarray*}\displaystyle \left(\sum_{i=1}^n \dfrac{a_i}{n} \right)^n \geq \displaystyle \prod_{i=1}^n a_i \end{eqnarray*}$$

Case 1. All the $$a_i$$ are equal.

Then

$$\left(\displaystyle \sum_{i=1}^n \dfrac {a_i}{n}\right)^n = \displaystyle \prod_{i=1}^n a_i =$$ (maximum product),

by AM-GM condition for equality. The sum $$\displaystyle \sum_{i=1}^n a_i$$ is fixed. The difference between any of the $$a_i$$
is zero or minimized since they are all equal; hence the difference between the $$a_i$$ can not be decreased. Since condition ii of Lemma is not satisfied, Lemma states that the product can not be increased while having the sum fixed.
Since by AM-GM the product is maximum, the product can not be increased while having the sum fixed. Hence, Lemma is true when all $$a_i$$ are equal.

Case 2. Not all the $$a_i$$ are equal.

By AM-GM condition for strict inequality,

$$\displaystyle \left(\sum_{i=1}^n \dfrac{a_i}{n} \right)^n > \displaystyle \prod_{i=1}^n a_i \Leftrightarrow \alpha^n &>& \displaystyle \prod_{i=1}^n a_i \Leftrightarrow \displaystyle \prod_{i=1}^n \alpha > \displaystyle \prod_{i=1}^n a_i$$

as required. Therefore Lemma is true when not all $$a_i$$ are equal.

Since Lemma is true when all $$a_i$$ are equal, true when not all $$a_i$$ are equal, and there is no other relevant case, Lemma is true.

For the Math people out there, I receive any comments on the proof or else, very gratefully, as I am relatively new to this interest...

Last edited: May 3, 2007
6. May 2, 2007

### tim_lou

that looks like an old putnam problem... is it? 'cause it looks very familiar.

Anyway, I spent some time on it and couldn't solve it. tried substitution of $b_i=a_i-\frac{\pi}{4}$ but no luck there.

BTW, is there anything in your last post that is related to the solution? plus the lemma you just posted... I suppose it is under the constrain that the sum is constant? since otherwise I can make the product of two number arbitrarily big by solely increasing one of the numbers. the two conditions do not have to be satisfied.

Last edited: May 2, 2007
7. May 2, 2007

### teleport

That's definitely true. I guess I wasn't careful enough explaining myself. Right now I'm in sort of a hurry, but I'll try post something more on the problem later. It's an old USAMO problem, but I have seen it posted on various places, so the concept of the problem must be important.

8. May 2, 2007

### tim_lou

yeah... I suppose smoothing would solve it.

using substitution of $x_i=\tan(a_i-\frac{\pi}{4})[/tex] the problem can be solved if one can find the minimum of $$\prod \frac{1+x_i}{1-x_i}$$ under the constrain of $$\sum x_i=const$$ which can be derived from the "smoothing" thing you mentioned by proving the following: $$\frac{1+x}{1-x}\cdot \frac{1+y}{1-y}\ge \left(\frac{1+\frac{x+y}{2}}{1-\frac{x+y}{2}}\right)^2$$ for [itex]x,y\in (0,1)$,

and I believed that it follows immediately from Jensen when taking natural log of both side.

from there one shows that the product of tan is minimized when the constrain is minimum, ie
$$\sum x_i=n-1$$

and also,
$$x_i=x_j=\frac{n-1}{n+1}$$

so
$$\prod \frac{1+x_i}{1-x_i}\ge n^{n+1}$$

Last edited: May 2, 2007
9. May 3, 2007

### teleport

Hey that's almost perfect :tongue:. However, maybe you weren't very careful when you said that part can be shown by Jensen. Try to look at the interval of convexity for the function $$f(x) = ln \dfrac{1 + x}{1 - x}$$. This is precisely why the solution needs to be 'smooth'. I will not show the solution because you are very close...

10. May 3, 2007

### tim_lou

I see what you mean, x could be negative, and that adds a layer of complication there. the inequality for the smoothing part doesn't even work when x and y are both negative (in fact, the inequality is flipped).

11. May 3, 2007

### tim_lou

I suppose one can patch things up by arguing that, if
$$\sum x_i\ge n-1$$
$$x_i\in(-1,1)$$

then at most 1 x_i can be negative, since otherwise, suppose two x_i or more are negative,
$$\sum x_i < n+1-2 =n-1$$ a contradiction.

so if all xi are positive, the proof is done.

suppose x_0 is negative,
then perhaps somethings can be derived from that.

12. May 3, 2007

### teleport

That's one way to do it. Now you have to show that

$$x_0 + x_i > 0$$

for

$$i = 1,..., n$$

and use that.

13. May 4, 2007

### tim_lou

yeah, it is indeed quite obvious that $x_0+x_i>0$

since otherwise you'll get $$n-1>\sum x_i$$

so, I guess it remains to prove,
for y>x (make x_0--> -x)
$$\frac{1-x}{1+x}\cdot\frac{1+y}{1-y}\ge \left(\frac{1+\frac{-x+y}{2}}{1-\frac{-x+y}{2}}\right)^2$$

expanding the sides, the inequality is equivalent to:
$$2(y^3-x^3)+2xy(y-x)\ge 0$$

which is obviously true since y>x

so smoothing still works, and the inequality still holds.

I see what you mean by "smoothing", if all x_i are positive, we wouldn't even need all that smoothing technique. Jensen alone will eat the problem alive :tongue2: and ymmm....

Last edited: May 4, 2007
14. May 4, 2007

### teleport

Haha, yes. But I think this approach is more economic of time: (in the case I wouldn't have to show the Lemma is true )
For some

$$t, p \in (-1,1)$$,

where

$$t \geq p$$

consider the product

$$\dfrac{t + 1}{1 - t} \dfrac{1 + p}{1 - p} = \dfrac{2t + 2p}{1 - t - p + tp} + 1$$

Notice the term in the denominator

$$tp$$

So by Lemma, if we increase tp, while having the sum t + p fixed (for the purpose of using the given)then we decrease the above product. Equivalently, the product

$$\prod \dfrac{1 + x_i}{1 - x_i}$$

can be decreased by performing a similar T as in the proof of Lemma. So you get

$$\prod \dfrac{1 + x_i}{1 - x_i} \geq \left(\dfrac{\dfrac{\sum x_i}{n + 1} + 1}{1 - \dfrac{\sum x_i}{n + 1}}\right)^{n + 1}$$

and you get the result from the given by using the fact that the function

$$f(x) = \dfrac{1 + x}{1 - x}$$

is increasing.

Also, instead of expanding like you did, you might do the trick of replacing

$$x_0 + x_i$$

with

$$\dfrac{x_0 + x_i}{2} + \dfrac{x_0 + x_i}{2}$$

and since now every 'x' is positive, you can use Jensen right away.

Nice problem ah!

BTW three in a day? What are you taking? Thank god I'm taking first year Eng Phys, :tongue: .

Last edited: May 4, 2007
15. May 4, 2007

### tim_lou

yeah, that is a nice problem. Before this thread, I have no idea what smoothing (though I've heard of it). After some time of googling, I can safely say I can almost master the technique.

BTW, I have all my math finals line up on one day, and all physics in another.... looks like that will be the trend in my future too...but I take it as a good thing, as I only have to go to school twice for all my finals (I'm a commuter). ;)

Last edited: May 4, 2007
16. May 5, 2007

### teleport

Yeah, was a good thread. Writing about it and discussing it is a great way to learn/understand. You also gave ideas I didn't even think about. I will be taking a vacation now, but in the future I'll try to find some other nice topic to post.