The arithmetic mean-geometric mean (AM-GM) inequality states that the arithmetic mean of non-negative real numbers is greater than or equal to the geometric mean of the same list. Further, equality holds if and only if every number in the list is the same. Mathematically, for a collection of non-negative real numbers , we have
with equality if and only if .
This wiki page will familiarize you with the AM-GM inequality and its applications in several scenarios. We will also prove this inequality through several methods and further generalize it for deeper insights.
The Arithmetic Mean - Geometric Mean inequality, or AM-GM inequality, states the following:
The geometric mean cannot exceed the arithmetic mean, and they will be equal if and only if all the chosen numbers are equal. That is,
with equality if and only if .
To get comfortable, let's consider the case when i.e. when there are only two variables, say and . Then the AM-GM inequality says
Cross multiplying and rearranging terms, we get or . This is true because squares are always non-negative. Also equality occurs when or .
The general case is slightly harder to show, as we cannot just cross multiply and manipulate. A common approach would be inducting on the number of variables. We state the proof by Cauchy below.
Here is a simple example based on the AM-GM inequality.
If the product of two positive numbers is , what is the minimum value of their sum?
Let the two numbers be and . We are given that , and want to find the minimum value of .
The AM-GM inequality states that
implying for this problem that
Note: This minimum is achieved when .
You can try a problem similar to the above example:
A jelly shop sells its products in two different sets: 3 red jelly cubes and 3 green jelly cuboids.
The 3 red cubes are of side lengths while the 3 green cuboids are identical with dimensions as shown above.
Which option would give you more jelly?
AM-GM inequality can be proved by several methods. Some of them are listed here.
The first one in the list is to prove by some sort of induction. Here we go:
At first, we let the inequality for variables be asserted by . Traditional inductions check a base case and then show that . But we'll show these:
Why should this work? Note that we jump from to by showing . Then using , we can induct backwards from to to verify that all numbers between and (inclusive) satisfy the assertion. This is known as forward-backward induction.
Now we move on to prove those points. It's already shown above that holds, and now we prove . Consider positive reals . Since we assume that is true, for any positive reals holds. Then
where means AM-GM inequality applied on variables. We've also used the base case, with variables. So the second part of our proof is also complete. It remains to show that . To show this, we take positive reals and . Then notice that
So we have
As you must've noticed, the first inequality here is variable AM-GM. So the third part of the proof is complete, and the induction as well. Clearly the equalities, inductively, hold if and only if .
We see that the AM-GM inequality is one of the special cases of Jensen's inequality. So let's approach it that way.
Consider the function
We observe that . Therefore, turns out to be a concave function. Also notice that we can conclude is concave using the graph of Thus, by Jensen's inequality, we have
By the property of logarithm, we have and . Therefore, we can simplify the terms on the RHS as
Another proof, made famous by a mathematician George Pólya, does not rely on induction. This shows us a more general version of the AM-GM inequality.
Suppose that for a sequence of positive reals with and a sequence of positive reals with such that , Pólya's proof begins with the observation that which can easily be verified graphically with equality occurring only at . This intuition allegedy came to Pólya in a dream, where he claimed it was "the best mathematics he had ever dreamt." If we make the change of variables , our initial observation becomes . Applying this to our sequence we get Now, we can see that However, we can also see from the latter of our initial observations that which is the same bound we just found for . So we could say So we have related and by inequality, but we have not separated them. Now we look closer at the case where and are equal to the expression on the right. The idea of "normalization" then comes to mind. In other words, we can somewhat manipulate the sequence to our advantage. Define a new sequence with where we have Applying our earlier bound for for our new variables , we get Then we have and it follows that So now, we are led to the fact that Since we have we can finally say
The following proof is way more intuitive and requires a bit of combinatorics.
Let be a set of positive real numbers. This set contains terms, and the power (which is a positive integer because represents the number of elements on a set) of this summation can be expressed as follows: We know that there exists a term which is the product of all the terms of . As we can take one term per bracket to combine with the other brackets, the combined term appears times, because there are ways to choose a term on the first bracket, on the second one, and so on.
As every term is positive, any power expansion is greater than its summand, and thus we get Now, we only need to remember the definition of geometric mean, which is the root of the product between all terms on a set followed by some algebraic manipulations. Then we achieve where is the geometric mean. Now we can prove by induction that for every positive integer , , and this proof will help us out to eliminate the factorial on the denominator of the last inequality. For this becomes , which is true. Then we assume the inequality is valid for and apply the inequality for , and prove that must be true too. Then, there we go: But from induction hypothesis, we have
With and , we get . But for we already saw that there's equality, and it becomes as we wished to prove. Now, we can use the inequality , which is valid for every positive integer. Some algebraic manipulations can be made as follows: where is the arithmetic mean. Now, a geometric interpretation on the real line makes the results easy to understand: When equals its limit, can only be smaller than . Otherwise, both equal each other. Hence, the inequality we were searching for:
Main Article: Applying AM-GM
You can refer to the article linked above for further problem solving on the applications of AM-GM inequality. In this section, we will work through some examples and problems based on the usage of AM-GM inequality.
If and , then find the minimum value of .
The given expression can be rewritten as
Therefore, the minimum value of is .
Note: The equality holds true when These values can be derived by noting that the minimum value is achieved when along with the given constraint that
Show that for .
We apply the 3-variable version of AM-GM with and to obtain
Then we multiply both sides by 3 to obtain .
Find all real solutions to .
We have , so . This implies is the only possible value. Since and , we have verified is the only solution.
Find all positive real solutions to
By AM-GM, we have . Summing these three inequalities, we obtain
Furthermore, summing the three given equations, we obtain
Hence, equality must hold throughout, implying and . By substituting these values into the original equations, we see that is indeed a solution.
[2-variable Cauchy Schwarz Inequality]
Solution 1: Expanding both sides, we can cancel terms and , so we need to show that . This follows from the 2-variable AM-GM by setting and , to obtain
Solution 2: From completing the square's Fermat's two square theorem, we have
Since squares are non-negative, the right-hand side is greater than or equal to .
Show that if , and are positive real numbers, then
A direct application of AM-GM doesn’t seem to work. Let's consider how we can get terms on the right hand side through AM-GM. To get , we will need more of than of or (as in the first example in this section). This gives a hint to try Similarily, we have and Adding these 3 inequalities and dividing by 4 yields
Let's work out the following example problems and try-it-yourself problem:
Find the minimum value of
We can rewrite the expression as
Since , we can find the minimum by using AM-GM:
So the minimum value is .
The weighted AM-GM inequality states that for non-negative numbers and non-negative weights
To prove the weighted AM-GM inequality, we will use the same approach of Jensen's inequality as we used above to prove AM-GM inequality. Here we will use the finite form of Jensen's inequality for the natural logarithm.
All values with weight have no influence on the inequality, so let's assume that all weights are positive. If all are equal, then equality holds. Therefore, it remains to prove strict inequality if they are not all equal. If at least one is zero (but not all), then the weighted geometric mean is zero, while the weighted arithmetic mean is positive, and hence strict inequality holds. Thus, assuming all are not equal, since the natural logarithm is strictly concave, the finite form of Jensen's inequality and the functional equations of the natural logarithm say
We know that the natural logarithm is strictly increasing, and therefore we have