Using mathematical induction

proof this formula.

the formula is attached the jpg files

[Sorry I hard to using type that formula so I captured that :)

Please help.T_T

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- Mar 30th 2010, 02:01 AMRelMathematical Induction Problem.
Using mathematical induction

proof this formula.

the formula is attached the jpg files

[Sorry I hard to using type that formula so I captured that :)

Please help.T_T - Mar 30th 2010, 02:43 AMsa-ri-ga-ma
For n = 2 you can prove it easily.

(a1+a2)/2 > sqrt(a1*a2)

(a1+a2) > 2*sqrt(a1*a2)

(a1+a2)^2 >4(a1*a2)

(a1+a2)^2 - 4(a1*a2) > 0

(a1-a2)^2 > 0 - Mar 30th 2010, 03:11 AMRel
- Mar 30th 2010, 08:09 AMArchie Meade
Hi Rel,

You could consider using the properties of logarithms.

**P(n)**

$\displaystyle \frac{a_1+a_2+a_3+....+a_n}{n}\ \ge\ \left(a_1\ a_2\ a_3\ ....a_n\right)^{\frac{1}{n}}$

We want to show that this causes P(n+1) to be true

**P(n+1)**

$\displaystyle \frac{a_1+a_2+a_3+...+a_n+a_{n+1}}{n+1}\ \ge\ \left(a_1\ a_2\ a_3\ ....a_n\ a_{n+1}\right)^{\frac{1}{n+1}}$

**Proof**

$\displaystyle \frac{a_1}{n}+\frac{a_2}{n}+\frac{a_3}{n}+...+\fra c{a_n}{n}\ \ge\ (a_1)^{\frac{1}{n}}\ (a_2)^{\frac{1}{n}}\ (a_3)^{\frac{1}{n}}\ ....(a_n)^{\frac{1}{n}}$

$\displaystyle \left(\frac{n}{n+1}\right)\left(\frac{a_1}{n}+\fra c{a_2}{n}+....+\frac{a_n}{n}+\frac{a_{n+1}}{n}\rig ht)\ \ge\ (a_1)^{\frac{1}{n+1}}\ (a_2)^{\frac{1}{n+1}}\ .....(a_n)^{\frac{1}{n+1}}\ (a_{n+1})^{\frac{1}{n+1}}$ ?

$\displaystyle \Rightarrow\ \frac{n}{n+1}\left[\frac{a_1}{n}+....+\frac{a_n}{n}\right]+\frac{a_{n+1}}{n+1}\ \ge\ (a_1)^{\frac{1}{n}\ \frac{n}{n+1}}\ (a_2)^{\frac{1}{n}\ \frac{n}{n+1}}....(a_n)^{\frac{1}{n}\ \frac{n}{n+1}}\ (a_{n+1})^{\frac{1}{n}\ \frac{n}{n+1}}$

$\displaystyle \Rightarrow\ \frac{n}{n+1}\left[\frac{a_1}{n}+....+\frac{a_n}{n}\right]+\frac{a_{n+1}}{n+1}\ \ge\ \left((a_1)^{\frac{1}{n}}\ (a_2)^{\frac{1}{n}}....(a_n)^{\frac{1}{n}}\right)^ {\frac{n}{n+1}}\ (a_{n+1})^{\frac{1}{n+1}}$

Now, notice that

$\displaystyle log\left[\left((a_1)^{\frac{1}{n}}\ (a_2)^{\frac{1}{n}}.....(a_n)^{\frac{1}{n}}\right) ^{\frac{n}{n+1}}\ (a_{n+1})^{\frac{1}{n+1}}\right]$ $\displaystyle =\frac{n}{n+1}\left((a_1)^{\frac{1}{n}}\ (a_2)^{\frac{1}{n}}...(a_n)^{\frac{1}{n}}\right)+\ frac{a_{n+1}}{n+1}$

and the proof follows since

$\displaystyle x>logx$

Whoops!! massive typo!! I left out the logs in the RHS

so there is no proof!! - Mar 30th 2010, 09:48 AMtonio
- Apr 2nd 2010, 09:44 AMArchie Meade
Thanks tonio,

starting from scratch,

we can apply a few manipulations for convenience...

If the following is true

$\displaystyle \frac{A_1+A_2+......+A_n}{n}\ \ge\ \sqrt[n]{A_1A_2....A_n}$

we attempt to prove that it causes

$\displaystyle \frac{A_1+A_2+....+A_n+A_{n+1}}{n+1}\ \ge\ \sqrt[n+1]{A_1A_2....A_nA_{n+1}}$

to be true.

Rewriting $\displaystyle a_i=\sqrt[n]{A_i}$ for i=1,2,....,n

we try to prove

$\displaystyle {a_1}^n+{a_2}^n+....+{a_n}^n\ \ge\ na_1a_2....a_n$

by showing that it causes

$\displaystyle {a_1}^{k+1}+{a_2}^{k+1}+...{a_k}^{k+1}+{a_{k+1}}^{ k+1}\ \ge\ (k+1)a_1a_2....a_ka_{k+1}$

**Proof**

$\displaystyle {a_1}^{k+1}+{a_2}^{k+1}+....+{a_k}^{k+1}+{a_{k+1}} ^{k+1}=a_1{a_1}^k+a_2{a_2}^k+....+\color{red}a_k\c olor{black}{a_k}^k+\color{blue}a_{k+1}\color{black }{a_{k+1}}^k$

To develop the inequality, the following may be utilised...

$\displaystyle x_1\ \ge\ x_2,\ y_1\ \ge\ y_2\ \Rightarrow\ x_1y_1+x_2y_2\ \ge\ x_1y_2+x_2y_1$

This is because

$\displaystyle x_1y_1+x_2y_2-(x_1y_2+x_2y_1)=x_1(y_1-y_2)-x_2(y_1-y_2)=(x_1-x_2)(y_1-y_2)\ \ge\ 0$

Hence, if the terms are arranged in decreasing order...

$\displaystyle a_1\ \ge\ a_2\ \ge\ a_3\ \ge.....\ge\ a_k\ \ge\ a_{k+1}$

$\displaystyle a_1{a_1}^k+a_2{a_2}^k+....+\color{red}a_k\color{bl ack}{a_k}^k+\color{blue}a_{k+1}\color{black}{a_{k+ 1}}^k\ \ge\ a_1{a_1}^k+a_2{a_2}^k+....+\color{blue}a_{k+1}\col or{black}{a_k}^k+\color{red}a_k\color{black}{a_{k+ 1}}^k$

Therefore

$\displaystyle a_1{a_1}^k+a_2{a_2}^k+...+a_k{a_k}^k+a_{k+1}{a_{k+ 1}}^k\ \ge\ (a_{k+1}{a_1}^k+a_{k+1}{a_2}^k+....a_{k+1}{a_k}^k) +a_k{a_{k+1}}^k$

$\displaystyle \ge\ a_{k+1}({a_1}^k+{a_2}^k+...+{a_k}^k)+a_k{a_{k+1}}^ k$

Hence

$\displaystyle {a_1}^{k+1}+{a_2}^{k+1}+....{a_k}^{k+1}+{a_{k+1}}^ {k+1}\ \ge\ a_{k+1}({a_1}^k+{a_2}^k+...+{a_k}^k)+a_k{a_{k+1}}^ k$

Since the terms are arranged in decreasing order, then $\displaystyle a_i\ \ge\ a_{k+1}$

Hence

$\displaystyle {a_1}^{k+1}+{a_2}^{k+1}+....{a_k}^{k+1}+{a_{k+1}}^ {k+1}\ \ge\ a_{k+1}({a_1}^k+{a_2}^k+...+{a_k}^k)+a_1a_2...a_ka _{k+1}$

$\displaystyle {a_1}^{k+1}+{a_2}^{k+1}+....{a_k}^{k+1}+{a_{k+1}}^ {k+1}\ \ge\ [a_{k+1}]\left[({a_1}^k+{a_2}^k+...+{a_k}^k)+a_1a_2...a_k\right]$

If the original statement is true, $\displaystyle {a_1}^k+{a_1}^k+...+{a_k}^k\ \ge\ ka_1a_2...a_k$

then

$\displaystyle {a_1}^{k+1}+{a_2}^{k+1}+....+{a_k}^{k+1}+{a_{k+1}} ^{k+1}\ \ge\ a_{k+1}[(ka_1a_2...a_k)+a_1a_2...a_k]$

$\displaystyle {a_1}^{k+1}+{a_2}^{k+1}+....+{a_k}^{k+1}+{a_{k+1}} ^{k+1}\ \ge\ (k+1)(a_1a_2....a_{k+1})$ - Jun 6th 2010, 04:16 PMMathematicalInductionAbove Answer does not work
Just to note, the above answer does not work. I'm not sure how to exactly solve it, but if you follow the above string of logic, it doesn't work for 2 reasons.

A) You state that "if we assume the sequence decreases" there is nothing in the original problem that says the sequence decreases, only that all numbers of a are greater than 0. they could all be positive and be completely random. It does not make a difference.

b) second of all, if one assumes that a1 is greater than a2 which is greater than a3 etc, then akak+1^k cannot be greater than a1a2....akak+1 simply because akak+1^k has the exact same amount of digits as a1a2.....akak+1, but since a1 is (by transitive property) greater than or equal to ak+1, then the inequality must be false (it can be equal to, but not greater than. If it was simply equal to, it would be a formula, not an inequality, so therefore, that one step does not connect with the previous one)

Other than that, the logic is fine, but since most of the problem only works ASSUMING that the sequence is decreasing, which the problem does not state. If you want to state that the sequence is decreasing, then you would need to prove it somehow. Also, if the sequence is decreasing, then that last chain of logic does not work, as stated above.

On the other hand, I'm not sure how to actually solve it, but I think I'm quite right in noticing the flaw(s) in your logic. I would like to see the actual answer to the problem, please reply!(Happy)

Oh, by the way, the above Mathematical Induction Problem is hardly Pre-Algebra to Algebra level. I don't even recall doing anything this hard when I did pre-calculus in high school. It should probably be moved to "University Math". Its also the reason why I joined, as I couldn't help but see the flaw, and I felt a need to correct it. I would really like to see a final result though, as I'm not quite sure how to go about solving it either. - Jun 6th 2010, 04:40 PMArchie Meade
- Jun 6th 2010, 04:48 PMMathematicalInduction
- Jun 6th 2010, 05:52 PMtonio
- Jun 6th 2010, 07:22 PMMathematicalInduction
- Jun 6th 2010, 08:06 PMsimplependulum

First show by induction if $\displaystyle a_1 a_2 ... a_n = 1 ~ a_i > 0$ then $\displaystyle a_1 + a_2 + ... + a_n \geq n $

When $\displaystyle n = 1 $ it is true by checking actually $\displaystyle a_1 = 1 \geq 1 $

Assume it is true for $\displaystyle a_1 ... a_k = 1 ~ \implies a_1 + ... + a_k \geq k $

When $\displaystyle n=k+1 $

Among these $\displaystyle k+1 $ numbers , let's choose $\displaystyle max\{\ a_i \}\ $ and $\displaystyle min\{\ a_i \}\$ wlog let them be $\displaystyle a_k $ and $\displaystyle a_{k+1} $ respectvely and we have $\displaystyle a_k \geq 1 $ and $\displaystyle a_{k+1} \leq 1 $

then $\displaystyle (a_k - 1 ) (a_{k+1} - 1 ) \leq 0 $

$\displaystyle \boxed{ a_k + a_{k+1} \geq 1 + a_k a_{k+1} }$

Then we have for $\displaystyle a_1 ... a_{k-1} (a_k a_{k+1} )= 1$

$\displaystyle a_1 + a_2 + ... + a_{k-1 } + (a_k a_{k+1}) \geq k $ (by asumption )

Thus

$\displaystyle a_1 + a_2 + ... + a_{k-1} + (a_k + a_{k+1}) \geq a_1 + a_2 + ... + a_{k-1} + (1 + a_k a_{k+1}) $

$\displaystyle = (a_1 + a_2 + ... + a_{k-1} + a_k a_{k+1} )+ 1 \geq k +1 $

Then let $\displaystyle a_i = \frac{b_i}{B}$ then we have

$\displaystyle b_1 b_2 ... b_n = B^n $ implies

$\displaystyle \frac{b_1 + b_2 + ... + b_n}{B} \geq n $

$\displaystyle \frac{b_1 + b_2 + ... + b_n}{n} \geq B = \sqrt[n]{ b_1 b_2 ... b_n } $

QED - Jun 7th 2010, 02:49 AMArchie Meade
- Jun 8th 2010, 06:27 PMMathematicalInduction
no problem. and thanks for solving it. it makes sense now.

- Jun 9th 2010, 11:22 AMFancyMouse
An easier way is to prove the n=2^k case by induction, and then for an arbitrary n, pad a[i] up to a power of 2 by the AM of a[1] to a[n]