Arithmetic Mean - Geometric Mean
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 \(n\) non-negative real numbers \(a_1,a_2,...,a_n\), we have
\[\frac{ a_1 + a_2 + \cdots + a_n } { n}\ge\sqrt[n] { a_1 a_2 \cdots a_n} , \]
with equality if and only if \(a_1=a_2=\cdots =a_n\).
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.
Contents
Statement of AM-GM inequality
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,
\[\frac{ a_1 + a_2 + \cdots + a_n } { n}\ge\sqrt[n] { a_1 a_2 \ldots a_n}\]
with equality if and only if \(a_1=a_2=\cdots =a_n\).
More precisely,
\[{\frac{\displaystyle \sum _{ i=1 }^{ n }{ { a }_{ i } }}{n} \ge \sqrt [ n ]{ \prod _{ i=1 }^{ n }{ { a }_{ i } } } }. \ _\square\]
To get comfortable, let's consider the case when \(n=2,\) i.e. when there are only two variables, say \(x\) and \(y\). Then the AM-GM inequality says
\[\dfrac{x+y}{2}\ge \sqrt{xy}.\]
Cross multiplying and rearranging terms, we get \(x-2\sqrt{xy}+y\ge 0\) or \(\left(\sqrt x-\sqrt y\right)^2\ge 0\). This is true because squares are always non-negative. Also equality occurs when \(\sqrt x=\sqrt y\) or \(x=y\).
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 \(2\) positive numbers is \(100\), what is the minimum value of their sum?
Let the \(2\) numbers be \(a\) and \(b\). We are given that \( ab = 100 \), and want to find the minimum value of \( a+ b \).
The AM-GM inequality states that
\[ \dfrac{ a + b } { 2 } \geq \sqrt{ab } ,\]
implying for this problem that \( a + b \geq 2 \sqrt{100} = 20 . \ _\square \)
Note: This minimum is achieved when \( a = b = 10 \).
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 \(a<b<c,\) while the 3 green cuboids are identical with dimensions \(a\times b\times c,\) as shown above.
Which option would give you more jelly?
Proof of AM-GM Inequality
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 \(n\) variables be asserted by \(P(n)\). Traditional inductions check a base case and then show that \(P(n)\implies P(n+1)\). But we'll show these:
- \(P(2)\) holds,
- \(P(n)\implies P(2n)\),
- \(P(n)\implies P(n-1)\).
Why should this work? Note that we jump from \(n\) to \(2n\) by showing \(P(n)\implies P(2n)\). Then using \(P(n)\implies P(n-1)\), we can induct backwards from \(2n\) to \(n+1\), to verify that all numbers between \(n\) and \(2n\) (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 \(P(2)\) holds, and now we prove \(P(n)\implies P(2n)\). Consider positive reals \(a_1,a_2,...,a_{2n}\). Since we assume that \(P(n)\) is true, for any \(n\) positive reals \(\frac{ \sum _{ i=1 }^{ n }{ { a }_{ i } }}{n} \ge \sqrt [ n ]{ \prod _{ i=1 }^{ n }{ { a }_{ i } } } \) holds. Then
\[\begin{eqnarray} \dfrac{1}{2n}\displaystyle\sum_{i=1}^{2n} a_i &=& \dfrac{1}{2n}\left(\displaystyle\sum_{i=1}^n a_i +\displaystyle \sum_{i=n+1}^{2n} a_i\right) \\ &\stackrel{n\text{ AM-GM}}{\ge} &\dfrac{1}{2}\left(\sqrt[n]{\prod_{i=1}^n a_i} + \sqrt[n]{\prod_{i=n+1}^{2n} a_i}\right) \\ &\stackrel{2\text{ AM-GM}}{\ge} &\sqrt{\sqrt[n]{\prod_{i=1}^n a_i} \times \sqrt[n]{\prod_{i=n+1}^{2n} a_i}} \\ &=& \sqrt[2n]{\prod_{i=1}^{2n} a_i}, \end{eqnarray}\]
where \(n\text{ AM-GM}\) means AM-GM inequality applied on \(n\) variables. We've also used the base case, with \(2\) variables. So the second part of our proof is also complete. It remains to show that \(P(n)\implies P(n-1)\). To show this, we take \(n\) positive reals \(a_1,a_2,...,a_{n-1}\) and \( a_n=\frac{1}{n-1} \sum_{i=1}^{n-1} a_i\). Then notice that \[\dfrac{1}{n}\sum_{i=1}^n a_i =\dfrac{1}{n}\left(\sum_{i=1}^{n-1} a_i + \dfrac{1}{n-1}\sum_{i=1}^{n-1} a_i\right)=\dfrac{1}{n-1}\sum_{i=1}^{n-1} a_i.\]
So we have
\[\begin{eqnarray} &~& \dfrac{1}{n-1}\displaystyle\sum_{i=1}^{n-1} a_i &=& \dfrac{1}{n}\sum_{i=1}^n a_i \\ &~& &\ge &\sqrt[n]{\prod_{i=1}^n a_i} \\ &~& &=& \sqrt[n]{\dfrac{1}{n-1}\left(\sum_{i=1}^{n-1} a_i\right) \left(\prod_{i=1}^{n-1} a_i\right)} \\ &\Rightarrow & \left(\dfrac{1}{n-1}\sum_{i=1}^{n-1}a_i \right)^n &\ge & \dfrac{1}{n-1}\left(\sum_{i=1}^{n-1} a_i\right) \left(\prod_{i=1}^{n-1} a_i\right) \\ &\Rightarrow & \left(\dfrac{1}{n-1}\sum_{i=1}^{n-1} a_i \right)^{n-1} &\ge & \prod_{i=1}^{n-1} a_i \\ &\Rightarrow & \dfrac{1}{n-1}\sum_{i=1}^{n-1} a_i &\ge & \sqrt[n-1]{\prod_{i=1}^{n-1} a_i}. \end{eqnarray}\]
As you must've noticed, the first inequality here is \(n\) 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 \(a_1=a_2=\cdots =a_n\). \(_\square\)
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 \(f(x) = \log x \ \forall x > 0.\)
We observe that \(f^{\prime\prime}(x) = \frac{-1}{x^2} < 0\). Therefore, \(f(x)\) turns out to be a concave function. \((\)Also notice that we can conclude \(f(x)\) is concave using the graph of \(\log x.)\) Thus, by Jensen's inequality, we have
\[\begin{align} f\left(\dfrac{\sum_{i=1}^n a_i}{n}\right) & \geq \dfrac{\sum_{i=1}^n f\left(a_i\right)}{n} \\ \Rightarrow \log \left(\dfrac{\sum_{i=1}^n a_i}{n}\right) & \geq \dfrac{\sum_{i=1}^n \log \left(a_i\right)}{n}. \end{align}\]
By the property of logarithm, we have \(\log x + \log y = \log xy\) and \(b\log a = \log a^b\). Therefore, we can simplify the terms on the RHS as
\[\begin{align} \log \left(\dfrac{\sum_{i=1}^n a_i}{n}\right) & \geq \dfrac{\sum_{i=1}^n \log a_i}{n} \\ & = \dfrac{\log a_1+\log a_2+ \cdots + \log a_n}{n} \\ & = \dfrac{\log \left(a_1 a_2 a_3 \cdots a_n\right)}{n} \\ & = \log \left(\prod_{i=1}^n a_i\right)^{1/n}. \end{align}\]
Thus,
\[\dfrac{ \sum_{i=1}^n a_i}{n} \geq \sqrt[n]{\prod_{i=1}^n a_i}.\ _\square\]
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 \(a_k\) with \(1\leq k\leq n\) and a sequence of positive reals \(p_k\) with \(1\leq k \leq n\) such that \( \sum_{k=1}^{n} p_k = 1\), \[a_1^{p_1}a_2^{p_2}a_3^{p_3}\cdots a_n^{p_n}\leq a_1p_1+a_2p_2+a_3p_3+\cdots+a_np_n.\] Pólya's proof begins with the observation that \(1+x\leq e^x,\) which can easily be verified graphically with equality occurring only at \(x=0\). 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 \(x\mapsto x-1\), our initial observation becomes \(x\leq e^{x-1}\). Applying this to our sequence \(a_k,\) we get \[\begin{align} a_k &\leq e^{a_k -1}\\ a_k^{p_k} &\leq e^{a_kp_k - p_k}. \end{align}\] Now, we can see that \[\begin{align} G=\prod_{k=1}^{n} a_k^{p_k} &\leq \exp \left(\sum_{k=1}^{n} a_kp_k - \sum_{k=1}^{n} p_k\right)\\ &\leq \exp \left(\sum_{k=1}^{n} a_kp_k - 1\right). \end{align}\] However, we can also see from the latter of our initial observations that \[A=\sum_{k=1}^{n} a_kp_k \leq \exp \left(\sum_{k=1}^{n} a_kp_k -1\right),\] which is the same bound we just found for \(G\). So we could say \[A, G \leq \exp \left (\sum_{k=1}^{n} a_kp_k -1 \right ).\] So we have related \(A\) and \(G\) by inequality, but we have not separated them. Now we look closer at the case where \(A\) and \(G\) 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 \(a_k\) to our advantage. Define a new sequence \(\alpha_k\) with \(1\leq k \leq n,\) where we have \[\begin{align} \alpha_k &= \frac{a_k}{A}\\ A&=a_1p_1+a_2p_2+a_3p_3+\cdots+a_np_n. \end{align}\] Applying our earlier bound for \(G\) for our new variables \(\alpha_k\), we get \[\begin{align} \prod_{k=1}^{n} \alpha_k^{p_k} &\leq \exp \left(\sum_{k=1}^{n} \alpha_kp_k-1\right)\\ \prod_{k=1}^{n} \left(\frac{a_k}{A}\right)^{p_k} &\leq \exp \left(\sum_{k=1}^{n} \frac{a_k}{A}p_k-1\right). \end{align}\] Then we have \[\sum_{k=1}^{n} \frac{a_k}{A}p_k = \frac{a_1p_1+a_2p_2+a_3p_3+\cdots+a_np_n}{A} = \frac{A}{A} = 1,\] and it follows that \[\exp \left(\sum_{k=1}^{n} \frac{a_k}{A}p_k-1\right) = 1.\] So now, we are led to the fact that \[\begin{align} \prod_{k=1}^{n} \left(\frac{a_k}{A}\right)^{p_k} &\leq 1\\ \frac{\prod_{k=1}^{n} a_k^{p_k}}{\prod_{k=1}^{n} A^{p_k}} &\leq 1. \end{align}\] Since we have \(\displaystyle \sum_{k=1}^{n} p_k = 1,\) we can finally say \[\begin{align} \frac{G}{A} &\leq 1\\ G &\leq A\\ a_1^{p_1}a_2^{p_2}a_3^{p_3}...a_n^{p_n} &\leq a_1p_1+a_2p_2+a_3p_3+\cdots+a_np_n.\ _\square \end{align}\]
The following proof is way more intuitive and requires a bit of combinatorics.
Let \(S\) be a set of positive real numbers. This set contains \(n\) terms, and the \(n^\text{th}\) power (which is a positive integer because \(n\) represents the number of elements on a set) of this summation can be expressed as follows: \[(a_1+a_2+\cdots+a_n)^{n}=\underbrace{(a_1+a_2+\cdots+a_n) (a_1+a_2+\cdots+a_n) \cdots (a_1+a_2+\cdots+a_n)}_{n \text{ times}}.\] We know that there exists a term which is the product of all the terms of \(S\). As we can take one term per bracket to combine with the other brackets, the combined term \(a_1 a_2 a_3 \ldots a_n\) appears \(n!\) times, because there are \(n\) ways to choose a term on the first bracket, \(n-1\) on the second one, and so on.
As every term is positive, any \(n^\text{th}\) power expansion is greater than its summand, and thus we get \[\left(\sum_{i=1}^n a_i \right)^{n} >n! \left(\prod_{i=1}^n a_i\right).\] Now, we only need to remember the definition of geometric mean, which is the \(n^\text{th}\) root of the product between all terms on a set followed by some algebraic manipulations. Then we achieve \[\frac{ \sum_{i=1}^n a_i}{(n!)^{\frac{1}{n}}}=k> m_g,\] where \(m_g\) is the geometric mean. Now we can prove by induction that for every positive integer \(n\), \(n^{n} \geq n!\), and this proof will help us out to eliminate the factorial on the denominator of the last inequality. For \(n=1\) this becomes \(1^{1} \geq 1!\), which is true. Then we assume the inequality is valid for \(n=k\) and apply the inequality for \(n=k+1\), and prove that must be true too. Then, there we go: \[\begin{align} (k+1)^{k+1} &=k^{k+1}+(k+1)\cdot k^{k}\cdot 1^{1}+\frac{k\cdot (k+1)}{2}\cdot k^{k-1}\cdot 1^{2}+\cdots +1^{k+1}\\ &>k^{k+1}+(k+1)\cdot k^{k}\\ &>k^{k}\cdot (2k+1). \end{align}\] But from induction hypothesis, we have \[\begin{align} k^{k} &\geq k! \\ k^{k}.(2k+1) &\geq k!.(2k+1) \\ k^{k}.(2k+1) &\geq k!.[k+(k+1)]\\ k^{k}\cdot (2k+1) &\geq k\cdot k!+(k+1)! \\ &>(k+1)! \\ \Rightarrow k^{k}\cdot(2k+1)&>(k+1)!. \end{align}\]
With \((k+1)^{k+1}>k^{k}\cdot (2k+1)\) and \(k^{k}\cdot (2k+1)>(k+1)!\), we get \((k+1)^{k+1} >(k+1)!\). But for \(k=0\) we already saw that there's equality, and it becomes \((k+1)^{k+1} \geq (k+1)!\) as we wished to prove. Now, we can use the inequality \(n^{n} \geq n!\), which is valid for every positive integer. Some algebraic manipulations can be made as follows: \[\begin{align} n^{n} &\geq n! \\ \frac{1}{n^{n}} &\leq \frac{1}{n!} \\ \frac{1}{n} &\leq \frac{1}{(n!)^{\frac{1}{n}}} \\ \frac{ \sum_{i=1}^n a_i}{n} &\leq \frac{ \sum_{i=1}^n a_i}{(n!)^{\frac{1}{n}}} \\ m_a &\leq \frac{ \sum_{i=1}^n a_i}{(n!)^{\frac{1}{n}}}=k, \end{align}\] where \(m_a\) is the arithmetic mean. Now, a geometric interpretation on the real line makes the results easy to understand: \[\begin{align} k>m_g>0 &\implies m_g \in (0, k) \\ k\geq m_a>0 &\implies m_a \in (0, k]. \end{align}\] When \(m_a\) equals its limit, \(m_g\) can only be smaller than \(m_g\). Otherwise, both equal each other. Hence, the inequality we were searching for: \[m_g \leq m_a.\ _\square\]
Examples with AM-GM inequality
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 \(x,y\in{\Bbb R^{+}}\) and \(x+y=8\), then find the minimum value of \(\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\).
The given expression can be rewritten as
\[\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\frac{1+x+y+xy}{xy}=\frac{9+xy}{xy}=\frac{9}{xy}+1.\]
From AM-GM
\[\begin{align} \frac{x+y}{2}&\ge\sqrt{xy}\\ 4&\ge\sqrt{xy}\\ 16&\ge xy\\ \frac{1}{16}&\le\frac{1}{xy}\\ \Rightarrow \frac{9}{xy}+1&\geq\frac{25}{16}. \end{align}\]
Therefore, the minimum value of \(\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\) is \(\frac{25}{16}\). \(_\square\)
Note: The equality holds true when \(x=y=4.\) These values can be derived by noting that the minimum value is achieved when \(xy=16,\) along with the given constraint that \(x + y = 8.\)
Show that \( 2a^3 + b^3 \geq 3a^2 b\) for \(a, b >0\).
We apply the 3-variable version of AM-GM with \( x_1= a^3, x_2 = a^3\) and \( x_3 = b^3\) to obtain
\[ \frac { a^3 + a^3 + b^3 } {3} \geq \sqrt[3]{a^3\cdot a^3 \cdot b^3} = a^2 b.\]
Then we multiply both sides by 3 to obtain \( a^3 + a^3 + b^3 \geq 3a^2 b\). \(_\square\)
Find all real solutions to \( 2^x + x^2 = 2 - \frac {1}{2^x}\).
We have \( 2 \leq 2^x + \frac {1}{2^x} = 2 - x^2\), so \( 0 \geq x^2\). This implies \( x=0\) is the only possible value. Since \( 2^0 + 0^2 = 1\) and \( 2 - \frac {1}{2^0} = 1\), we have verified \( x=0\) is the only solution. \(_\square\)
Find all positive real solutions to
\[ \begin{array} &4x + \frac {18}{y} = 14, &2y + \frac {9}{z} = 15, &9z + \frac {16}{x} = 17. \end{array} \]
By AM-GM, we have \( 4x + \frac {16}{x} \geq 16, 2y + \frac {18}{y} \geq 12, 9z + \frac {9}{z} \geq 18\). Summing these three inequalities, we obtain
\[ 4x + \frac {16}{x} + 2y + \frac {18}{y} + 9z + \frac {9}{z} \geq 16 + 12 + 18 = 46 .\]
Furthermore, summing the three given equations, we obtain
\[ 4x + \frac {16}{x} + 2y + \frac {18}{y} + 9z + \frac {9}{z} = 46.\]
Hence, equality must hold throughout, implying \( x=2, y=3\) and \( z=1\). By substituting these values into the original equations, we see that \( (x, y, z) = (2, 3, 1)\) is indeed a solution. \(_\square\)
[2-variable Cauchy Schwarz Inequality]
Show \(\left(a^2 + b^2\right)\left(c^2 + d^2\right) \geq (ac+bd)^2 .\)
Solution 1: Expanding both sides, we can cancel terms \( a^2c^2\) and \( b^2d^2\), so we need to show that \( a^2 d^2 + b^2c^2 \geq 2acbd\). This follows from the 2-variable AM-GM by setting \( x_1 = a^2 d^2\) and \( x_2 = b^2 c^2\), to obtain
\[ a^2d^2 + b^2c^2 \geq 2 \lvert abcd \rvert \geq 2 abcd.\]
Solution 2: From completing the square's Fermat's two square theorem, we have
\[ \big(a^2+b^2\big)\big(c^2+d^2\big) = (ac+bd)^2 + (ad-bc)^2 .\]
Since squares are non-negative, the right-hand side is greater than or equal to \( (ac+bd)^2\). \(_\square\)
Show that if \( a, b\), and \( c\) are positive real numbers, then
\[ a^4 + b^4 + c^4 \geq abc(a+b+c). \]
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 \( a^2bc\), we will need more of \( a\) than of \( b\) or \( c\) (as in the first example in this section). This gives a hint to try \[ a^4 + a^4 + b^4 + c^4 \geq 4 a^2 b c.\] Similarily, we have \[ a^4 +b^4 +b^4 +c^4 \geq 4ab^2 c\] and \[ a^4 + b^4 + c^4 + c^4 \geq 4abc^2.\] Adding these 3 inequalities and dividing by 4 yields \[ a^4 + b^4 + c^4 \geq a^2bc + ab^2c + abc^2 = abc(a+b+c).\ _\square\]
Let's work out the following example problems and try-it-yourself problem:
Find the minimum value of
\[ \frac{ 4+ 9x^2\sin^2 x}{x\sin x} ~\text{ for }~ 0 < x <\pi.\]
We can rewrite the expression \(\frac{4+9x^{2}\sin{x}^{2}}{x \sin{x}}\) as
\[\frac{4}{x \sin{x}}+9x\sin{x}.\]
Since \(x>0\), we can find the minimum by using AM-GM:
\[\begin{align} \frac{\frac{4}{x \sin{x}}+9x\sin{x}}{2}&\geq\sqrt{36}\\ \frac{4}{x \sin{x}}+9x\sin{x}&\geq12. \end{align}\]
So the minimum value is \(12\). \(_\square\)
Generalization: Weighted AM-GM Inequality
The weighted AM-GM inequality states that for non-negative numbers \( a_1,...,a_n \) and non-negative weights \( \omega_1,...,\omega_n \)
\[ \large\dfrac {\displaystyle\sum \omega_i a_i } { \displaystyle\sum \omega_i } \geq \sqrt[\sum\omega_i]{\displaystyle\prod a_i ^ {\omega_i }}.\ _\square\]
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 \(a_k\) with weight \(\omega_k = 0\) have no influence on the inequality, so let's assume that all weights are positive. If all \(a_k\) are equal, then equality holds. Therefore, it remains to prove strict inequality if they are not all equal. If at least one \(a_k\) 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 \(a_k\) 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
\[ \begin{align} \ln\Bigl(\frac{\omega_1a_1+\cdots+\omega_na_n} \omega \Bigr) & >\frac{w_1}w\ln a_1+\cdots+\frac{w_n}w\ln x_n \\ & =\ln \left( \sqrt[\sum\omega_i]{a_1^{\omega_1} x_2^{\omega_2} \cdots a_n^{\omega_n}} \right) . \end{align} \]
We know that the natural logarithm is strictly increasing, and therefore we have
\[\frac{\omega_1a_1+\cdots+\omega_na_n}{\omega} > \sqrt[\sum\omega_i]{a_1^{\omega_1} a_2^{\omega_2} \cdots a_n^{\omega_n}}.\ _\square\]