Difference between revisions of "Proofs of AM-GM"

(started page; more proofs to come)
 
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=== Proof by Convexity ===
 
=== Proof by Convexity ===
  
We note that the [[function]] <math>x \mapsto \ln x</math> is trictly [[concave]].  Then by [[Jensen's Inequality]],
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We note that the [[function]] <math>x \mapsto \ln x</math> is strictly [[concave]].  Then by [[Jensen's Inequality]],
 
<cmath> \ln \sum_i \lambda_i a_i \ge \lambda_i \sum_i \ln a_i = ln \prod_i a_i^{\lambda_i} , </cmath>
 
<cmath> \ln \sum_i \lambda_i a_i \ge \lambda_i \sum_i \ln a_i = ln \prod_i a_i^{\lambda_i} , </cmath>
 
with equality if and only if all the <math>a_i</math> are equal.
 
with equality if and only if all the <math>a_i</math> are equal.
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<cmath> \sum_i \lambda_i a_i \ge \prod_i a_i^{\lambda_i}, </cmath>
 
<cmath> \sum_i \lambda_i a_i \ge \prod_i a_i^{\lambda_i}, </cmath>
 
with equality if and only if all the <math>a_i</math> are equal, as desired.  <math>\blacksquare</math>
 
with equality if and only if all the <math>a_i</math> are equal, as desired.  <math>\blacksquare</math>
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=== Alternate Proof by Convexity ===
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''This proof is due to [[G. Pólya]].''
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Note that the function <math>f:x \mapsto e^x</math> is strictly convex.  Let <math>g(x)</math> be the line tangent to <math>f</math> at <math>(0,1)</math>; then <math>g(x) = x+1</math>.  Since <math>f</math> is also a [[continuous]], [[differentiable]] function, it follows that <math>f(x) \ge g(x)</math> for all <math>x</math>, with equality exactly when <math>x=0</math>, i.e.,
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<cmath> 1+x \le e^x, </cmath>
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with equality exactly when <math>x=0</math>.
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Now, set
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<cmath> r_i = a_i/\biggl( \sum_{j=1}^n \lambda_j a_j \biggr) - 1, </cmath>
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for all integers <math>1\le i \le n</math>.  Our earlier bound tells us that
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<cmath> r_i +1 \le \exp(r_i), </cmath>
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so
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<cmath> (r_i +1)^{\lambda_i} \le \exp(\lambda_i r_i) .</cmath>
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Multiplying <math>n</math> such inequalities gives us
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<cmath> \frac{\prod_{i=1}^n a_i^{\lambda_i}}{ \sum_{i=1}^n \lambda_i r_i} = \prod_{i=1}^n (r_i + 1) \le \prod_{i=1}^n \exp \lambda_i r_i = \exp \sum_{i=1}^n \lambda_i a_i / \sum_{j=1}^n \lambda_j a_j - \lambda_i = \exp 0 = 1 ,</cmath>
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so
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<cmath> \prod_{i=1}^n a_i^{\lambda_i} \le \sum_{i=1}^n \lambda_i r_i , </cmath>
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as desired.  Equality holds when <math>r_i</math> is always equal to zero, i.e., when <math>a_i = a_j</math> for all <math>i,j</math> such that <math>\lambda_i , \lambda_j \neq 0</math>.  <math>\blacksquare</math>
  
 
== Proofs of Unweighted AM-GM ==
 
== Proofs of Unweighted AM-GM ==
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Define the <math>n</math> sequence <math>\{ r_{ij}\}_{i=1}^{n}</math> as <math>r_{ij} = \sqrt[n]{a_i}</math>, for all integers <math>1 \le i,j \le n</math>.  Evidently these sequences are similarly sorted.  Then by the [[Rearrangement Inequality]],
 
Define the <math>n</math> sequence <math>\{ r_{ij}\}_{i=1}^{n}</math> as <math>r_{ij} = \sqrt[n]{a_i}</math>, for all integers <math>1 \le i,j \le n</math>.  Evidently these sequences are similarly sorted.  Then by the [[Rearrangement Inequality]],
<cmath> \sum_i a_i = \sum_i \prod_j r_{ij} \ge \sum_i \prod_j r(i,i+j) = n \prod_i \sqrt[n]{a_i} , </cmath>
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<cmath> \sum_i a_i = \sum_i \prod_j r_{ij} \ge \sum_i \prod_j r_{i,i+j} = n \prod_i \sqrt[n]{a_i} , </cmath>
 
where we take our indices modulo <math>n</math>, with equality exactly when all the <math>r_{ij}</math>, and therefore all the <math>a_i</math>, are equal.  Dividing both sides by <math>n</math> gives the desired inequality.  <math>\blacksquare</math>
 
where we take our indices modulo <math>n</math>, with equality exactly when all the <math>r_{ij}</math>, and therefore all the <math>a_i</math>, are equal.  Dividing both sides by <math>n</math> gives the desired inequality.  <math>\blacksquare</math>
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=== Proof by Cauchy Induction ===
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We first prove that the inequality holds for two variables.  Note that <math>(\sqrt{a}, \sqrt{b}), (\sqrt{a}, \sqrt{b})</math> are similarly sorted sequences.  Then by the Rearrangement Inequality,
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<cmath> \sqrt{ab} = \frac{ \sqrt{a} \sqrt{b} + \sqrt{b} \sqrt{a}}{2} \le \frac{ \sqrt{a} \sqrt{a} + \sqrt{b} \sqrt{b}}{2} = \frac{a+b}{2} , </cmath>
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with equality exactly when <math>a=b</math>.
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We next prove that if the inequality holds for <math>n</math> variables (with equality when all are equal to zero), than it holds for <math>2n</math> variables (with equality when all are equal to zero).  Indeed, suppose the inequality holds for <math>n</math> variables.  Let <math>A_n, A'_n, A_{2n}</math> denote the arithmetic means of <math>(a_1, \dotsc, a_n)</math>, <math>(a_{n+1}, \dotsc, a_{2n})</math>, <math>(a_1, \dotsc, a_{2n})</math>, respectively; let <math>G_n, G'_n, G_{2n}</math> denote their respective geometric means.  Then
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<cmath> G_{2n} = \sqrt{G_n G'_n} \le \frac{G_n + G'_n}{2} \le \frac{A_n + A'_n}{2} = A_2n, </cmath>
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with equality when all the numbers <math>a_1, \dotsc, a_{2n}</math> are equal, as desired.
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These two results show by induction that the theorem holds for <math>n=2^k</math>, for all integers <math>k</math>.  In particular, for every integer <math>n</math>, there is an integer <math>N >n</math> such that the theorem holds for <math>N</math> variables.
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Finally, we show that if <math>N>n</math> and the theorem holds for <math>N</math> variables, then it holds for <math>n</math> variables <math>a_1, \dotsc, a_n</math> with arithmetic mean <math>A_n</math> and geometric mean <math>G_n</math>.  Indeed, set <math>b_k = a_k</math> for <math>1\le k \le n</math>, and let <math>b_k = A_k</math> for <math>n < k \le N</math>.  Then
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<cmath> G_k^{n/N} \cdot A_k^{(N-n)/N} = \prod_{i=1}^N b_i^{1/N} \le \sum_{i=1}^N b_i/N = A_n . </cmath>
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It follows that <math>G_k^{n/N} \le A_k^{n/N}</math>, or <math>G_n \le A_n</math>, with equality exactly when all the <math>a_i</math> are equal to <math>A_k</math>.
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The last two results show that for all positive integers <math>n</math>, the theorem holds for <math>n</math> variables.  Therefore the theorem is true.  <math>\blacksquare</math>
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[[Category:Inequality]]

Revision as of 21:50, 31 December 2007

This pages lists some proofs of the weighted AM-GM Inequality. The inequality's statement is as follows: for all nonnegative reals $a_1, \dotsc, a_n$ and nonnegative reals $\lambda_1, \dotsc, \lambda_n$ such that $\sum_{i=1}^n \lambda_i = 1$, then \[\sum_{i=1}^n \lambda_i a_i \ge \prod_{i=1}^n a_i^{\lambda_i},\] with equality if and only if $a_i = a_j$ for all $i,j$ such that $\lambda_i, \lambda_j \neq 0$.

We first note that we may disregard any $a_i$ for which $\lambda_i= 0$, as they contribute to neither side of the desired inequality. We also note that if $a_i= 0$ and $\lambda_i \neq 0$, for some $i$, then the right-hand side of the inequality is zero and the right hand of the inequality is greater or equal to zero, with equality if and only if $a_j = 0 = a_i$ whenever $\lambda_j\neq 0$. Thus we may henceforth assume that all $a_i$ and $\lambda_i$ are strictly positive.

Complete Proofs

Proof by Convexity

We note that the function $x \mapsto \ln x$ is strictly concave. Then by Jensen's Inequality, \[\ln \sum_i \lambda_i a_i \ge \lambda_i \sum_i \ln a_i = ln \prod_i a_i^{\lambda_i} ,\] with equality if and only if all the $a_i$ are equal. Since $x \mapsto \ln x$ is a strictly increasing function, it then follows that \[\sum_i \lambda_i a_i \ge \prod_i a_i^{\lambda_i},\] with equality if and only if all the $a_i$ are equal, as desired. $\blacksquare$

Alternate Proof by Convexity

This proof is due to G. Pólya.

Note that the function $f:x \mapsto e^x$ is strictly convex. Let $g(x)$ be the line tangent to $f$ at $(0,1)$; then $g(x) = x+1$. Since $f$ is also a continuous, differentiable function, it follows that $f(x) \ge g(x)$ for all $x$, with equality exactly when $x=0$, i.e., \[1+x \le e^x,\] with equality exactly when $x=0$.

Now, set \[r_i = a_i/\biggl( \sum_{j=1}^n \lambda_j a_j \biggr) - 1,\] for all integers $1\le i \le n$. Our earlier bound tells us that \[r_i +1 \le \exp(r_i),\] so \[(r_i +1)^{\lambda_i} \le \exp(\lambda_i r_i) .\] Multiplying $n$ such inequalities gives us \[\frac{\prod_{i=1}^n a_i^{\lambda_i}}{ \sum_{i=1}^n \lambda_i r_i} = \prod_{i=1}^n (r_i + 1) \le \prod_{i=1}^n \exp \lambda_i r_i = \exp \sum_{i=1}^n \lambda_i a_i / \sum_{j=1}^n \lambda_j a_j - \lambda_i = \exp 0 = 1 ,\] so \[\prod_{i=1}^n a_i^{\lambda_i} \le \sum_{i=1}^n \lambda_i r_i ,\] as desired. Equality holds when $r_i$ is always equal to zero, i.e., when $a_i = a_j$ for all $i,j$ such that $\lambda_i , \lambda_j \neq 0$. $\blacksquare$

Proofs of Unweighted AM-GM

These proofs use the assumption that $\lambda_i = 1/n$, for all integers $1 \le i \le n$.

Proof by Rearrangement

Define the $n$ sequence $\{ r_{ij}\}_{i=1}^{n}$ as $r_{ij} = \sqrt[n]{a_i}$, for all integers $1 \le i,j \le n$. Evidently these sequences are similarly sorted. Then by the Rearrangement Inequality, \[\sum_i a_i = \sum_i \prod_j r_{ij} \ge \sum_i \prod_j r_{i,i+j} = n \prod_i \sqrt[n]{a_i} ,\] where we take our indices modulo $n$, with equality exactly when all the $r_{ij}$, and therefore all the $a_i$, are equal. Dividing both sides by $n$ gives the desired inequality. $\blacksquare$

Proof by Cauchy Induction

We first prove that the inequality holds for two variables. Note that $(\sqrt{a}, \sqrt{b}), (\sqrt{a}, \sqrt{b})$ are similarly sorted sequences. Then by the Rearrangement Inequality, \[\sqrt{ab} = \frac{ \sqrt{a} \sqrt{b} + \sqrt{b} \sqrt{a}}{2} \le \frac{ \sqrt{a} \sqrt{a} + \sqrt{b} \sqrt{b}}{2} = \frac{a+b}{2} ,\] with equality exactly when $a=b$.

We next prove that if the inequality holds for $n$ variables (with equality when all are equal to zero), than it holds for $2n$ variables (with equality when all are equal to zero). Indeed, suppose the inequality holds for $n$ variables. Let $A_n, A'_n, A_{2n}$ denote the arithmetic means of $(a_1, \dotsc, a_n)$, $(a_{n+1}, \dotsc, a_{2n})$, $(a_1, \dotsc, a_{2n})$, respectively; let $G_n, G'_n, G_{2n}$ denote their respective geometric means. Then \[G_{2n} = \sqrt{G_n G'_n} \le \frac{G_n + G'_n}{2} \le \frac{A_n + A'_n}{2} = A_2n,\] with equality when all the numbers $a_1, \dotsc, a_{2n}$ are equal, as desired.

These two results show by induction that the theorem holds for $n=2^k$, for all integers $k$. In particular, for every integer $n$, there is an integer $N >n$ such that the theorem holds for $N$ variables.

Finally, we show that if $N>n$ and the theorem holds for $N$ variables, then it holds for $n$ variables $a_1, \dotsc, a_n$ with arithmetic mean $A_n$ and geometric mean $G_n$. Indeed, set $b_k = a_k$ for $1\le k \le n$, and let $b_k = A_k$ for $n < k \le N$. Then \[G_k^{n/N} \cdot A_k^{(N-n)/N} = \prod_{i=1}^N b_i^{1/N} \le \sum_{i=1}^N b_i/N = A_n .\] It follows that $G_k^{n/N} \le A_k^{n/N}$, or $G_n \le A_n$, with equality exactly when all the $a_i$ are equal to $A_k$.

The last two results show that for all positive integers $n$, the theorem holds for $n$ variables. Therefore the theorem is true. $\blacksquare$