Wikipedia:Reference desk/Archives/Mathematics/2009 July 14

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July 14

Proving inequalities

Say $a+b+c=1$. Due to the AM-GM inequality, $abc\leq {\frac {1}{27}}$, which means that the maximum value of abc is 1/27. Now, what values of a, b and c give this maximum value? Obviously it's when they're all 1/3, but is there some way to prove this? On a related topic, if I want to prove ${\sqrt {a+b}}+{\sqrt {a+c}}+{\sqrt {b+c}}\leq 6$ where $a+b+c=6$, is it fair to say that $3{\sqrt {a+b}}\leq 6$ (since you can swap the variables around...?) and somehow prove the inequality using this? Is there a name for the kinds of things I'm talking about, where you have inequalities with permutations of variables? --wj32 ^t/c 11:27, 14 July 2009 (UTC)

The AM-GM inequality says that the AM and GM are equal if and only if all terms are equal, which in your $a+b+c=1$ example gives $a=b=c=1/3$. For the second question, what you say isn't true in general - e.g. $a=6,b=c=0$. AndrewWTaylor (talk) 11:38, 14 July 2009 (UTC)

You can only "swap variables around" when there's symmetry. Suppose you start with the assumption $a+b+c=6$. So far a, b and c are symmetrical. Then you evaluate your sum, starting with ${\sqrt {a+b}}$. After this step, a and b are still symmetrical, but c is not symmetrical with them. You have included a summand which has a and b, but not one that contains c. So from this point forth, ${\sqrt {a+c}}$ is very different from ${\sqrt {a+b}}$ and you can't interchange them.

By the way, in your first question you have forgotten to state that $a,b,c\geq 0$ (not sure if this is required for the second). -- Meni Rosenfeld (talk) 12:55, 14 July 2009 (UTC)

My favorite way to see the AGM inequality is like this. Say x=abc where $a\neq b$. Now let $u={a+b \over 2}$ and $v={a-b \over 2}$. Suppose you adjust a and b while keeping their sum constant. That means that u doesn't change, while the product $x=(u+v)(u-v)c=(u^{2}-v^{2})c$ is clearly at maximum when v=0 which means a=b. By symmetry (since you could have permuted the variables in any way) the product is at maximum when a=b=c. —Preceding unsigned comment added by 70.90.174.101 (talk) 12:24, 14 July 2009 (UTC)

Lagrange multipliers shows the general technique for this sort of problem. HTH, Robinh (talk) 19:45, 14 July 2009 (UTC)

Thanks for all your answers. Say I changed the problem to that of proving ${\sqrt {\frac {a+b}{c}}}+{\sqrt {\frac {b+c}{a}}}+{\sqrt {\frac {c+a}{b}}}\leq x$ where $a+b+c=y$ and $a,b,c>0$. Could I use symmetry to prove this inequality, rewriting it as $3\cdot {\sqrt {\frac {a+b}{c}}}\leq x$? --wj32 ^t/c 01:20, 15 July 2009 (UTC)

No, are you sure you have read my explanation above?

Symmetry allows you to do things like assuming, without loss of generality, an ordering on the variables (say $a\leq b\leq c$). Since both the condition and the result are, as a whole, symmetrical, and there must be some ordering, you may as well assume that a is the least and c is the greatest. But you can't just change the inequality arbitrarily.

The whole point in such theorems is that to satisfy the condition, if one variable increases, another must decrease. Thus an increase in one summand is offset by a decrease in another. You destroy all that if you keep only one summand - you can change its value by playing with the variables without any consequences.

Anyway, neither of these statements can be valid, since the LHS is unbounded under the condition. -- Meni Rosenfeld (talk) 10:02, 15 July 2009 (UTC)

regarding Bernoulli trials

Dear Wikipedians:

The question reads "Bernoulli trials with probability p of success and q of failure, what is the probability of 3rd success on 7th trial and 5th success on 12th trial?"

I reasoned that 3rd success on 7th trial means exactly 2 successes in first 6 trials, and 5th success on 12th trial means exactly 1 success among the 8th, 9th, 10th and 11th trials (4 trials in total), therefore my answer to this question is:

(6 choose 2)p²q⁴ + (4 choose 1)pq³.

How sound is my reasoning?

Thanks for the help.

70.31.152.197 (talk) 16:46, 14 July 2009 (UTC)

First, 3rd success on the 7th trial means exactly 2 successes in the first 6 trials and a success on the 7th trial, so it's ${\binom {6}{2}}p^{3}q^{4}$. Similarly you need a success on the 12th trial, so the second part is ${\binom {4}{1}}p^{2}q^{3}$. Finally, you want both to happen, so you need to multiply the parts rather than add them. So it should be ${\binom {6}{2}}p^{3}q^{4}\cdot {\binom {4}{1}}p^{2}q^{3}=60p^{5}q^{7}$. -- Meni Rosenfeld (talk) 17:02, 14 July 2009 (UTC)

Simplifying this sum

Is there a simple expression for this sum?:

$\sum _{n=1}^{N}{\frac {1}{n^{2}}},\quad N\in \{1,2,\ldots \}$

—Bromskloss (talk) 20:46, 14 July 2009 (UTC)

(pi^2)/6 --pma (talk) 20:56, 14 July 2009 (UTC)

That's the limit as N tends to infinity. I don't think there's a nice expression for the partial sums. Algebraist 21:00, 14 July 2009 (UTC)

Ops, too much in a hurry, did'n notice the "N". However, we can easily write an analytic function S(z) such that the sum above is S(N). Can be of any interest? --pma (talk) 21:07, 14 July 2009 (UTC)

Maple gives ${\frac {\pi ^{2}}{6}}-\Psi (1,N+1)$ where $\Psi (n,x)$ is the nth Polygamma function. Whether or not you consider that simple is up to you. It's not elementary so I wouldn't consider it simple. Irish Souffle (talk) 21:05, 14 July 2009 (UTC)

If you want to exhibit your sum as a value of an analytic function at z=N, just write

$S(z)=\sum _{1}^{\infty }{\frac {1}{k^{2}}}-{\frac {1}{(k+z)^{2}}}$.

You can further expand in a geometric series each term and rearrange into a power series; you can do it by the absolute convergence (however, within a radius of convergence 1).This way in fact you find the quoted polygamma. Added note: the above S(x), restricted to the positive semi-axis, is the unique increasing solution to the functional equation S(x)=S(x-1)+ 1/x² for all x>1. --pma (talk) 21:28, 14 July 2009 (UTC)

PS: Irish Souffle, I don't agree with yours "not elementary hence not simple"; there are elementary things that are not at all simple and simple things that are not at all elementary; in fact usually in math we make things less elementary exactly in order to make them simpler --pma (talk) 21:34, 14 July 2009 (UTC)

It's simple in the sense that it is nice, neat, and compact, but if he was planning on calculating things by hand I'm not sure that he'd think of it as simple (unless there's an easy way to evaluate the polygamma function at integers; I've never used it before). Irish Souffle (talk) 21:46, 14 July 2009 (UTC)

It's also called a generalized harmonic number, specifically $H_{N}^{(2)}$. The Euler–Maclaurin formula can probably be used to evaluate it numerically. -- Meni Rosenfeld (talk) 22:09, 14 July 2009 (UTC)