វិសមភាព AM-GM: សម្រាយបញ្ជាក់នៃវិសមភាព AM–GM

សម្រាយបញ្ជាក់នៃវិសមភាព AM–GM 

យើងមានរបៀបក្នុងការស្រាយបញ្ជាក់វិសមភាព AM–GM នេះ; ឧទាហរណ៍ដូចជា, វាអាចស្រាយបញ្ជាក់ដោយប្រើប្រាស់ វិសមភាព Jensen, ដោយប្រើប្រាស់នូវអនុគមន៍ផត ln()។ យើងក៏អាចស្រាយបញ្ជាក់វាតាម វិសមភាពតំរៀប។ ផ្តោតទៅលើប្រវែងនិងតម្រូវការ, ការស្រាយបញ្ជាក់តាមកំនើនវាចារ ខាងក្រោមប្រហែលជា ជាការស្រាយបញ្ជាក់ដ៍ល្អមយួៈ

គំនិតនៃការស្រាយបញ្ជាក់ដំបូងទាំងពីរ

យើងត្រូវតែស្រាយបញ្ជាក់ថាៈ

ដែលសមភាពកើតឡើងលុះត្រាតែគ្រប់ចំនួនទាំងអស់ស្មើគ្នា។ បើ , យើងជំនួស  និង  ដោយ  នឹងធ្វើឲ្យមធ្យមនព្វន្តដែលនៅខាងឆ្វេងដៃ មិនមានការប្រែប្រួល តែវានឹងធ្វើឲ្យមធ្យមធរណីមាត្រដែលនៅខាងស្ដាំដៃមានការកើនឡើង ព្រោះ ៖

ដូចនេះអង្គខាងស្ដាំនឹង ធំជាងគេ — ដូចនេះគំនិតគឺថា — នៅពេលដែលគ្រប់  ទាំងអស់មានតម្លៃស្មើទៅនឹងមធ្យមនព្វន្ត ៖

ដូចនេះ ខាងក្រោមនេះគឺជាតម្លៃធំជាងគេនៃអង្គខាងស្ដាំ គឺ៖

ការស្រាយបញ្ជាក់នេះគឺពិតចំពោះករណី  តែដំណើរការក្នុងការយកមធ្យមនៃគូរចំនួនដែលស្រដៀងគ្នា អាចឈានទៅរកភាពបរាជ័យក្នុងការបង្កើតនូវ   ចំនួនដែលស្មើគ្នា ក្នុងករណី  ។ ជាឧទាហរណ៍មួយក្នុងករណីនេះ គឺ  : ដោយរកមធ្យមនៃចំនួនពីរដែលខុសគ្នាអាចនឹងបង្កើតបាន ចំនួនពីរដែលមានតម្លៃស្មើគ្នា ប៉ុន្តែ ចំនួនទី បី គឺនៅតែខុសគ្នាដដែល។ ដូចនេះ តាមពិតទៅ យើងមិនដែលទទួលបាននៅ វិសមភាពដែលមាន មធ្យមធរណីមាត្រដែលកើតពី ចំនួន៣ ដែលស្មើគ្នា នោះទេ។

ដូច្នេះ ល្បិចមួយទៀតដែលត្រូវប្រើនោះ គឺ ត្រូវតែបង្វែរ គំនិតខាងលើនេះឲ្យទៅជា ការស្រាយបញ្ជាក់ដ៏មានប្រសិទ្ធិ៍ភាពមួយ ចំពោះករណី  ។

Hard proof by induction

With the arithmetic mean

of the non-negative real numbers , the AM–GM statement is equivalent to

with equality if and only if  for all }.

For the following proof we apply mathematical induction and only well-known rules of arithmetic.

Induction basis:  For  the statement is true with equality.

Induction hypothesis: Suppose that the AM–GM statement holds for all choices of  non-negative real numbers.

Induction step: Consider  non-negative real numbers. Their arithmetic mean  satisfies

If all numbers are equal to , then we have equality in the AM–GM statement and we are done. Otherwise we may find one number that is greater than  and one that is smaller than , say  and  . Then

Now consider the  numbers  with

which are also non-negative. Since 

 is also the arithmetic mean of  numbers  and the induction hypothesis implies

Due to (*) we know that

hence

in particular . Therefore, if at least one of the numbers  is zero, then we already have strict inequality in (**). Otherwise the right-hand side of (**) is positive and strict inequality is obtained by using the estimate (***) to get a lower bound of the right-hand side of (**). Thus, in both cases we get

which completes the proof.

 Proof by Cauchy using forward-backward-induction

The following proof by cases relies directly on well-known rules of arithmetic but employs the rarely used technique of forward-backward-induction. It is essentially from Augustin Louis Cauchy and can be found in his Cours d'analyse.Cauchy, Augustin-Louis (1821). Cours d'analyse de l'École Royale Polytechnique, première partie, Analyse algébrique,  Paris. The proof of the inequality of arithmetic and geometric means can be found on pages 457ff.

 The case where all the terms are equal 

If all the terms are equal:

then their sum is , so their arithmetic mean is ; and their product is , so their geometric mean is ; therefore, the arithmetic mean and geometric mean are equal, as desired.

 The case where not all the terms are equal 

It remains to show that if not all the terms are equal, then the arithmetic mean is greater than the geometric mean. Clearly, this is only possible when .

This case is significantly more complex, and we divide it into subcases.

 The subcase where n = 2 

If , then we have two terms,  and , and since (by our assumption) not all terms are equal, we have:

hence

as desired.

 The subcase where n = 2k 

Consider the case where , where  is a positive integer. We proceed by mathematical induction.

In the base case, , so . We have already shown that the inequality holds when , so we are done.

Now, suppose that for a given , we have already shown that the inequality holds for , and we wish to show that it holds for . To do so, we apply the inequality twice for  numbers and once for  numbers to obtain:

 

where in the first inequality, the two sides are equal only if

and

(in which case the first arithmetic mean and first geometric mean are both equal to , and similarly with the second arithmetic mean and second geometric mean); and in the second inequality, the two sides are only equal if the two geometric means are equal. Since not all  numbers are equal, it is not possible for both inequalities to be equalities, so we know that:

as desired.

 The subcase where n < 2k 

If  is not a natural power of , then it is certainly less than some natural power of 2, since the sequence  is unbounded above. Therefore, without loss of generality, let  be some natural power of  that is greater than .

So, if we have  terms, then let us denote their arithmetic mean by , and expand our list of terms thus:

We then have:

 

so

and 

as desired.

Proof by induction using basic calculus

The following proof uses mathematical induction and some basic differential calculus. 

Induction  basis: For  the statement is true with equality.

Induction hypothesis: Suppose that the AM–GM statement holds for all choices of  non-negative real numbers.

Induction step: In order to prove the statement for  non-negative real numbers , we need to prove that

with equality only if all the  numbers are equal.

If all numbers are zero, the inequality holds with equality. If some but not all numbers are zero, we have strict inequality. Therefore, we may assume in the following, that all  numbers are positive.

We consider the last number  as a variable and define the function

Proving the induction step is equivalent to showing that  for all , with  only if  and  are all equal. This can be done by analyzing the critical points of  using some basic calculus.

The first derivative of  is given by

A critical point  has to satisfy , which means

After a small rearrangement we get

and finally

which is the geometric mean of . This is the only critical point of . Since  for all , the function  is strictly convex and has a strict global minimum at . Next we compute the value of the function at this global minimum:

where the final inequality holds due to the induction hypothesis. The hypothesis also says that we can have equality only when  are all equal. In this case, their geometric mean   has the same value, Hence, unless  are all equal, we have . This completes the proof.

This technique can be used in the same manner to prove the generalized AM–GM inequality and Cauchy–Schwarz inequality in Euclidean space .

Proof by Pólya using the exponential function

George Pólya provided a proof similar to what follows. Let  for all real , with first derivative  and second derivative . Observe that ,  and  for all real , hence  is strictly convex with the absolute minimum at . Hence  for all real  with equality only for . 

Consider a list of non-negative real numbers . If they are all zero, then the AM–GM inequality holds with equality. Hence we may assume in the following for their arithmetic mean . By -fold application of the above inequality, we obtain that

with equality if and only if  for every . The argument of the exponential function can be simplified:

Returning to ,

which produces , hence the result