Example 1

If ${\displaystyle a,b,c>0}$, then the A.M.-G.M. tells us that

${\displaystyle (1+a)(1+b)(1+c)\geq 2{\sqrt {1\cdot {a}}}\cdot 2{\sqrt {1\cdot {b}}}\cdot 2{\sqrt {1\cdot {c}}}=8{\sqrt {abc}}}$

Example 2

A simple upper bound for ${\displaystyle n!}$ can be found. AM-GM tells us

${\displaystyle 1+2+\dots +n\geq n{\sqrt[{n}]{n!}}}$

${\displaystyle {\frac {n(n+1)}{2}}\geq n{\sqrt[{n}]{n!}}}$

and so

${\displaystyle \left({\frac {n+1}{2}}\right)^{n}\geq n!}$

with equality at ${\displaystyle n=1}$.

Equivalently,

${\displaystyle (n+1)^{n}\geq 2^{n}n!}$

Example 3

Consider the function

${\displaystyle f(x,y,z)={\frac {x}{y}}+{\sqrt {\frac {y}{z}}}+{\sqrt[{3}]{\frac {z}{x}}}}$

for all positive real numbers x, y and z. Suppose we wish to find the minimal value of this function. It can be rewritten as:

${\displaystyle {\begin{aligned}f(x,y,z)&=6\cdot {\frac {{\frac {x}{y}}+{\frac {1}{2}}{\sqrt {\frac {y}{z}}}+{\frac {1}{2}}{\sqrt {\frac {y}{z}}}+{\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}+{\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}+{\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}}{6}}\\&=6\cdot {\frac {x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}}\end{aligned}}}$

with

${\displaystyle x_{1}={\frac {x}{y}},\qquad x_{2}=x_{3}={\frac {1}{2}}{\sqrt {\frac {y}{z}}},\qquad x_{4}=x_{5}=x_{6}={\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}.}$

Applying the AM–GM inequality for n = 6, we get

${\displaystyle {\begin{aligned}f(x,y,z)&\geq 6\cdot {\sqrt[{6}]{{\frac {x}{y}}\cdot {\frac {1}{2}}{\sqrt {\frac {y}{z}}}\cdot {\frac {1}{2}}{\sqrt {\frac {y}{z}}}\cdot {\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}\cdot {\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}\cdot {\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}}}\\&=6\cdot {\sqrt[{6}]{{\frac {1}{2\cdot 2\cdot 3\cdot 3\cdot 3}}{\frac {x}{y}}{\frac {y}{z}}{\frac {z}{x}}}}\\&=2^{2/3}\cdot 3^{1/2}.\end{aligned}}}$

Further, we know that the two sides are equal exactly when all the terms of the mean are equal:

${\displaystyle f(x,y,z)=2^{2/3}\cdot 3^{1/2}\quad {\mbox{when}}\quad {\frac {x}{y}}={\frac {1}{2}}{\sqrt {\frac {y}{z}}}={\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}.}$

All the points (x, y, z) satisfying these conditions lie on a half-line starting at the origin and are given by

${\displaystyle (x,y,z)={\biggr (}t,{\sqrt[{3}]{2}}{\sqrt {3}}\,t,{\frac {3{\sqrt {3}}}{2}}\,t{\biggr )}\quad {\mbox{with}}\quad t>0.}$

Proof using Jensen's inequality

Jensen's inequality states that the value of a concave function of an arithmetic mean is greater than or equal to the arithmetic mean of the function's values. Since the logarithm function is concave, we have

${\displaystyle \log \left({\frac {\sum _{i}x_{i}}{n}}\right)\geq \sum {\frac {1}{n}}\log x_{i}=\sum \left(\log x_{i}^{1/n}\right)=\log \left(\prod x_{i}^{1/n}\right).}$

Taking antilogs of the far left and far right sides, we have the AM–GM inequality.

Proof by successive replacement of elements

We have to show that

${\displaystyle \alpha ={\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}=\beta }$

with equality only when all numbers are equal.

If not all numbers are equal, then there exist ${\displaystyle x_{i},x_{j}}$ such that ${\displaystyle x_{i}<\alpha <x_{j}}$. Replacing x_i by ${\displaystyle \alpha }$ and x_j by ${\displaystyle (x_{i}+x_{j}-\alpha )}$ will leave the arithmetic mean of the numbers unchanged, but will increase the geometric mean because

${\displaystyle \alpha (x_{j}+x_{i}-\alpha )-x_{i}x_{j}=(\alpha -x_{i})(x_{j}-\alpha )>0}$

If the numbers are still not equal, we continue replacing numbers as above. After at most ${\displaystyle (n-1)}$ such replacement steps all the numbers will have been replaced with ${\displaystyle \alpha }$ while the geometric mean strictly increases at each step. After the last step, the geometric mean will be ${\displaystyle {\sqrt[{n}]{\alpha \alpha \cdots \alpha }}=\alpha }$, proving the inequality.

It may be noted that the replacement strategy works just as well from the right hand side. If any of the numbers is 0 then so will the geometric mean thus proving the inequality trivially. Therefore we may suppose that all the numbers are positive. If they are not all equal, then there exist ${\displaystyle x_{i},x_{j}}$ such that ${\displaystyle 0<x_{i}<\beta <x_{j}}$. Replacing ${\displaystyle x_{i}}$ by ${\displaystyle \beta }$ and ${\displaystyle x_{j}}$ by ${\displaystyle {\frac {x_{i}x_{j}}{\beta }}}$leaves the geometric mean unchanged but strictly decreases the arithmetic mean since

${\displaystyle x_{i}+x_{j}-\beta -{\frac {x_{i}x_{j}}{\beta }}={\frac {(\beta -x_{i})(x_{j}-\beta )}{\beta }}>0}$. The proof then follows along similar lines as in the earlier replacement.

Induction proofs

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.^[4]

Proof by induction using basic calculus

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

Induction basis: For n = 1 the statement is true with equality.

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

Induction step: In order to prove the statement for n + 1 non-negative real numbers x₁, . . . , x_n, x_n+1, we need to prove that

${\displaystyle {\frac {x_{1}+\cdots +x_{n}+x_{n+1}}{n+1}}-({x_{1}\cdots x_{n}x_{n+1}})^{\frac {1}{n+1}}\geq 0}$

with equality only if all the n + 1 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 n + 1 numbers are positive.

We consider the last number x_n+1 as a variable and define the function

${\displaystyle f(t)={\frac {x_{1}+\cdots +x_{n}+t}{n+1}}-({x_{1}\cdots x_{n}t})^{\frac {1}{n+1}},\qquad t>0.}$

Proving the induction step is equivalent to showing that f(t) ≥ 0 for all t > 0, with f(t) = 0 only if x₁, . . . , x_n and t are all equal. This can be done by analyzing the critical points of f using some basic calculus.

The first derivative of f is given by

${\displaystyle f'(t)={\frac {1}{n+1}}-{\frac {1}{n+1}}({x_{1}\cdots x_{n}})^{\frac {1}{n+1}}t^{-{\frac {n}{n+1}}},\qquad t>0.}$

A critical point t₀ has to satisfy f′(t₀) = 0, which means

${\displaystyle ({x_{1}\cdots x_{n}})^{\frac {1}{n+1}}t_{0}^{-{\frac {n}{n+1}}}=1.}$

After a small rearrangement we get

${\displaystyle t_{0}^{\frac {n}{n+1}}=({x_{1}\cdots x_{n}})^{\frac {1}{n+1}},}$

and finally

${\displaystyle t_{0}=({x_{1}\cdots x_{n}})^{\frac {1}{n}},}$

which is the geometric mean of x₁, . . . , x_n. This is the only critical point of f. Since f′′(t) > 0 for all t > 0, the function f is strictly convex and has a strict global minimum at t₀. Next we compute the value of the function at this global minimum:

${\displaystyle {\begin{aligned}f(t_{0})&={\frac {x_{1}+\cdots +x_{n}+({x_{1}\cdots x_{n}})^{1/n}}{n+1}}-({x_{1}\cdots x_{n}})^{\frac {1}{n+1}}({x_{1}\cdots x_{n}})^{\frac {1}{n(n+1)}}\\&={\frac {x_{1}+\cdots +x_{n}}{n+1}}+{\frac {1}{n+1}}({x_{1}\cdots x_{n}})^{\frac {1}{n}}-({x_{1}\cdots x_{n}})^{\frac {1}{n}}\\&={\frac {x_{1}+\cdots +x_{n}}{n+1}}-{\frac {n}{n+1}}({x_{1}\cdots x_{n}})^{\frac {1}{n}}\\&={\frac {n}{n+1}}{\Bigl (}{\frac {x_{1}+\cdots +x_{n}}{n}}-({x_{1}\cdots x_{n}})^{\frac {1}{n}}{\Bigr )}\\&\geq 0,\end{aligned}}}$

where the final inequality holds due to the induction hypothesis. The hypothesis also says that we can have equality only when x₁, . . . , x_n are all equal. In this case, their geometric mean  t₀ has the same value, Hence, unless x₁, . . . , x_n, x_n+1 are all equal, we have f(x_n+1) > 0. 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 Rⁿ.

Proof by Pólya using the exponential function

George Pólya provided a proof similar to what follows. Let f(x) = e^x–1 – x for all real x, with first derivative f′(x) = e^x–1 – 1 and second derivative f′′(x) = e^x–1. Observe that f(1) = 0, f′(1) = 0 and f′′(x) > 0 for all real x, hence f is strictly convex with the absolute minimum at x = 1. Hence x ≤ e^x–1 for all real x with equality only for x = 1.

Consider a list of non-negative real numbers x₁, x₂, . . . , x_n. If they are all zero, then the AM–GM inequality holds with equality. Hence we may assume in the following for their arithmetic mean α > 0. By n-fold application of the above inequality, we obtain that

${\displaystyle {\begin{aligned}{{\frac {x_{1}}{\alpha }}{\frac {x_{2}}{\alpha }}\cdots {\frac {x_{n}}{\alpha }}}&\leq {e^{{\frac {x_{1}}{\alpha }}-1}e^{{\frac {x_{2}}{\alpha }}-1}\cdots e^{{\frac {x_{n}}{\alpha }}-1}}\\&=\exp {\Bigl (}{\frac {x_{1}}{\alpha }}-1+{\frac {x_{2}}{\alpha }}-1+\cdots +{\frac {x_{n}}{\alpha }}-1{\Bigr )},\qquad (*)\end{aligned}}}$

with equality if and only if x_i = α for every i ∈ {1, . . . , n}. The argument of the exponential function can be simplified:

${\displaystyle {\begin{aligned}{\frac {x_{1}}{\alpha }}-1+{\frac {x_{2}}{\alpha }}-1+\cdots +{\frac {x_{n}}{\alpha }}-1&={\frac {x_{1}+x_{2}+\cdots +x_{n}}{\alpha }}-n\\&={\frac {n\alpha }{\alpha }}-n\\&=0.\end{aligned}}}$

Returning to (*),

${\displaystyle {\frac {x_{1}x_{2}\cdots x_{n}}{\alpha ^{n}}}\leq e^{0}=1,}$

which produces x₁ x₂ · · · x_n ≤ αⁿ, hence the result^[5]

${\displaystyle {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}\leq \alpha .}$

Proof by Lagrangian multipliers

If any of the ${\displaystyle x_{i}}$ are ${\displaystyle 0}$, then there is nothing to prove. So we may assume all the ${\displaystyle x_{i}}$ are strictly positive.

Because the arithmetic and geometric means are homogeneous of degree 1, without loss of generality assume that ${\displaystyle \prod _{i=1}^{n}x_{i}=1}$. Set ${\displaystyle G(x_{1},x_{2},\ldots ,x_{n})=\prod _{i=1}^{n}x_{i}}$, and ${\displaystyle F(x_{1},x_{2},\ldots ,x_{n})={\frac {1}{n}}\sum _{i=1}^{n}x_{i}}$. The inequality will be proved (together with the equality case) if we can show that the minimum of ${\displaystyle F(x_{1},x_{2},...,x_{n}),}$ subject to the constraint ${\displaystyle G(x_{1},x_{2},\ldots ,x_{n})=1,}$ is equal to ${\displaystyle 1}$, and the minimum is only achieved when ${\displaystyle x_{1}=x_{2}=\cdots =x_{n}=1}$. Let us first show that the constrained minimization problem has a global minimum.

Set ${\displaystyle K=\{(x_{1},x_{2},\ldots ,x_{n})\colon 0\leq x_{1},x_{2},\ldots ,x_{n}\leq n\}}$. Since the intersection ${\displaystyle K\cap \{G=1\}}$ is compact, the extreme value theorem guarantees that the minimum of ${\displaystyle F(x_{1},x_{2},...,x_{n})}$ subject to the constraints ${\displaystyle G(x_{1},x_{2},\ldots ,x_{n})=1}$ and ${\displaystyle (x_{1},x_{2},\ldots ,x_{n})\in K}$ is attained at some point inside ${\displaystyle K}$. On the other hand, observe that if any of the ${\displaystyle x_{i}>n}$, then ${\displaystyle F(x_{1},x_{2},\ldots ,x_{n})>1}$, while ${\displaystyle F(1,1,\ldots ,1)=1}$, and ${\displaystyle (1,1,\ldots ,1)\in K\cap \{G=1\}}$. This means that the minimum inside ${\displaystyle K\cap \{G=1\}}$ is in fact a global minimum, since the value of ${\displaystyle F}$ at any point inside ${\displaystyle K\cap \{G=1\}}$ is certainly no smaller than the minimum, and the value of ${\displaystyle F}$ at any point ${\displaystyle (y_{1},y_{2},\ldots ,y_{n})}$ not inside ${\displaystyle K}$ is strictly bigger than the value at ${\displaystyle (1,1,\ldots ,1)}$, which is no smaller than the minimum.

The method of Lagrange multipliers says that the global minimum is attained at a point ${\displaystyle (x_{1},x_{2},\ldots ,x_{n})}$ where the gradient of ${\displaystyle F(x_{1},x_{2},\ldots ,x_{n})}$ is ${\displaystyle \lambda }$ times the gradient of ${\displaystyle G(x_{1},x_{2},\ldots ,x_{n})}$, for some ${\displaystyle \lambda }$. We will show that the only point at which this happens is when ${\displaystyle x_{1}=x_{2}=\cdots =x_{n}=1}$ and ${\displaystyle F(x_{1},x_{2},...,x_{n})=1.}$

Compute ${\displaystyle {\frac {\partial F}{\partial x_{i}}}={\frac {1}{n}}}$ and

${\displaystyle {\frac {\partial G}{\partial x_{i}}}=\prod _{j\neq i}x_{j}={\frac {G(x_{1},x_{2},\ldots ,x_{n})}{x_{i}}}={\frac {1}{x_{i}}}}$

along the constraint. Setting the gradients proportional to one another therefore gives for each ${\displaystyle i}$ that ${\displaystyle {\frac {1}{n}}={\frac {\lambda }{x_{i}}},}$ and so ${\displaystyle n\lambda =x_{i}.}$ Since the left-hand side does not depend on ${\displaystyle i}$, it follows that ${\displaystyle x_{1}=x_{2}=\cdots =x_{n}}$, and since ${\displaystyle G(x_{1},x_{2},\ldots ,x_{n})=1}$, it follows that ${\displaystyle x_{1}=x_{2}=\cdots =x_{n}=1}$ and ${\displaystyle F(x_{1},x_{2},\ldots ,x_{n})=1}$, as desired.

Weighted AM–GM inequality

There is a similar inequality for the weighted arithmetic mean and weighted geometric mean. Specifically, let the nonnegative numbers x₁, x₂, . . . , x_n and the nonnegative weights w₁, w₂, . . . , w_n be given. Set w = w₁ + w₂ + · · · + w_n. If w > 0, then the inequality

${\displaystyle {\frac {w_{1}x_{1}+w_{2}x_{2}+\cdots +w_{n}x_{n}}{w}}\geq {\sqrt[{w}]{x_{1}^{w_{1}}x_{2}^{w_{2}}\cdots x_{n}^{w_{n}}}}}$

holds with equality if and only if all the x_k with w_k > 0 are equal. Here the convention 0⁰ = 1 is used.

If all w_k = 1, this reduces to the above inequality of arithmetic and geometric means.

One stronger version of this, which also gives strengthened version of the unweighted version, is due to Aldaz. In particular, There is a similar inequality for the weighted arithmetic mean and weighted geometric mean. Specifically, let the nonnegative numbers x₁, x₂, . . . , x_n and the nonnegative weights w₁, w₂, . . . , w_n be given. Assume further that the sum of the weights is 1. Then ${\displaystyle \sum _{i=1}^{n}w_{i}x_{i}\geq \prod _{i=1}^{n}x_{i}^{w_{i}}+\sum _{i=1}^{n}w_{i}\left(x_{i}^{\frac {1}{2}}-\sum _{i=1}^{n}w_{i}x_{i}^{\frac {1}{2}}\right)^{2}}$. ^[6]

Matrix arithmetic–geometric mean inequality

Most matrix generalizations of the arithmetic geometric mean inequality apply on the level of unitarily invariant norms, owing to the fact that even if the matrices ${\displaystyle A}$ and ${\displaystyle B}$ are positive semi-definite the matrix ${\displaystyle AB}$ may not be positive semi-definite and hence may not have a canonical square root. In ^[7] Bhatia and Kittaneh proved that for any unitarily invariant norm ${\displaystyle |||\cdot |||}$ and positive semi-definite matrices ${\displaystyle A}$ and ${\displaystyle B}$ it is the case that

${\displaystyle |||AB|||\leq {\frac {1}{2}}|||A^{2}+B^{2}|||}$

Later, in ^[8] the same authors proved the stronger inequality that

${\displaystyle |||AB|||\leq {\frac {1}{4}}|||(A+B)^{2}|||}$

Finally, it is known for dimension ${\displaystyle n=2}$ that the following strongest possible matrix generalization of the arithmetic-geometric mean inequality holds, and it is conjectured to hold for all ${\displaystyle n}$

${\displaystyle |||(AB)^{\frac {1}{2}}|||\leq {\frac {1}{2}}|||A+B|||}$

This conjectured inequality was shown by Stephen Drury in 2012. Indeed, he proved^[9]

${\displaystyle {\sqrt {\sigma _{j}(AB)}}\leq {\frac {1}{2}}\lambda _{j}(A+B),\ j=1,\ldots ,n.}$

Finance: Link to Geometric Asset Returns

In finance much research is concerned with accurately estimating the rate of return of an asset over multiple periods in the future. In the case of lognormal asset returns, there is an exact formula to compute the arithmetic asset return from the geometric asset return.

For simplicity, assume we are looking at yearly geometric returns r₁, r₂, ... , r_N over a time horizon of N years, i.e.

${\displaystyle r_{n}={\frac {V_{n}-V_{n-1}}{V_{n-1}}},}$

where:

${\displaystyle V_{n}}$ = value of the asset at time ${\displaystyle n}$,

${\displaystyle V_{n-1}}$ = value of the asset at time ${\displaystyle n-1}$.

The geometric and arithmetic returns are respectively defined as

${\displaystyle g_{N}=\left(\prod _{n=1}^{N}(1+r_{n})\right)^{1/N},}$

${\displaystyle a_{N}={\frac {1}{N}}\sum _{n=1}^{N}r_{n}.}$

When the yearly geometric asset returns are lognormally distributed, then the following formula can be used to convert the geometric average return to the arithemtic average return:^[10]

${\displaystyle 1+g_{N}={\frac {1+a_{N}}{\sqrt {1+{\frac {\sigma ^{2}}{(1+a_{N})^{2}}}}}},}$

where ${\displaystyle \sigma ^{2}}$ is the variance of the observed asset returns This implicit equation for a_N can be solved exactly as follows. First, notice that by setting

${\displaystyle z=(1+a_{N})^{2},}$

we obtain a polynomial equation of degree 2:

${\displaystyle z^{2}-(1+g)^{2}-(1+g)^{2}\sigma ^{2}=0.}$

Solving this equation for z and using the definition of z, we obtain 4 possible solutions for a_N:

${\displaystyle a_{N}=\pm {\frac {1+g_{N}}{\sqrt {2}}}{\sqrt {1\pm {\sqrt {1+{\frac {4\sigma ^{2}}{(1+g_{N})^{2}}}}}}}-1.}$

However, notice that

${\displaystyle {\sqrt {1+{\frac {4\sigma ^{2}}{(1+g_{N})^{2}}}}}\geq 1.}$

This implies that the only 2 possible solutions are (as asset returns are real numbers):

${\displaystyle a_{N}=\pm {\frac {1+g_{N}}{\sqrt {2}}}{\sqrt {1+{\sqrt {1+{\frac {4\sigma ^{2}}{(1+g_{N})^{2}}}}}}}-1.}$

Finally, we expect the derivative of a_N with respect to g_N to be non-negative as an increase in the geometric return should never cause a decrease in the arithmetic return. Indeed, both measure the average growth of an asset's value and therefore should move in similar directions. This leaves us with one solution to the implicit equation for a_N, namely

${\displaystyle a_{N}={\frac {1+g_{N}}{\sqrt {2}}}{\sqrt {1+{\sqrt {1+{\frac {4\sigma ^{2}}{(1+g_{N})^{2}}}}}}}-1.}$

Therefore, under the assumption of lognormally distributed asset returns, the arithmetic asset return is fully determined by the geometric asset return.

Other generalizations

Geometric proof without words that max (a,b) > root mean square (RMS) or quadratic mean (QM) > arithmetic mean (AM) > geometric mean (GM) > harmonic mean (HM) > min (a,b) of two distinct positive numbers a and b ^{[note 1]}

Other generalizations of the inequality of arithmetic and geometric means include:

• Muirhead's inequality,
• Maclaurin's inequality,
• Generalized mean inequality,
• Means of complex numbers.^[11]