# Difference between revisions of "Proofs of AM-GM"

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− | <cmath> \prod_{i=1}^n \exp \lambda_i r_i = \prod_{i=1}^n \exp (\frac{a_i \lambda_i}{ \sum_{i=1}^n \lambda_j a_j} - \lambda_i) = \exp (\frac{\sum_{i=1}^n a_i \lambda_i}{ \sum_{i=1}^n \lambda_j a_j} - \sum_{i=1}^n \lambda_i) = \exp 0 = 1 .</cmath> | + | <cmath> \prod_{i=1}^n \exp \lambda_i r_i = \prod_{i=1}^n \exp \left(\frac{a_i \lambda_i}{ \sum_{i=1}^n \lambda_j a_j} - \lambda_i\right) = \exp \left(\frac{\sum_{i=1}^n a_i \lambda_i}{ \sum_{i=1}^n \lambda_j a_j} - \sum_{i=1}^n \lambda_i\right) = \exp 0 = 1 .</cmath> |

Substituting the results for the left and right sides: | Substituting the results for the left and right sides: |

## Revision as of 08:47, 21 July 2021

This pages lists some proofs of the weighted AM-GM Inequality. The inequality's statement is as follows: for all nonnegative reals and nonnegative reals such that , then with equality if and only if for all such that .

We first note that we may disregard any for which , as they contribute to neither side of the desired inequality. We also note that if and , for some , then the right-hand side of the inequality is zero and the left hand of the inequality is greater or equal to zero, with equality if and only if whenever . Thus we may henceforth assume that all and are *strictly positive.*

## Contents

## Complete Proofs

### Proof by Convexity

We note that the function is strictly concave. Then by Jensen's Inequality, with equality if and only if all the are equal. Since is a strictly increasing function, it then follows that with equality if and only if all the are equal, as desired.

### Alternate Proof by Convexity

*This proof is due to G. Pólya.*

Note that the function is strictly convex. Let be the line tangent to at ; then . Since is also a continuous, differentiable function, it follows that for all , with equality exactly when , i.e., with equality exactly when .

Now, set for all integers . Our earlier bound tells us that so Multiplying such inequalities gives us

Evaluating the left hand side:

for

Evaluating the right hand side:

Substituting the results for the left and right sides:

as desired.

## Proofs of Unweighted AM-GM

These proofs use the assumption that , for all integers .

### Proof by Rearrangement

Define the sequence as , for all integers . Evidently these sequences are similarly sorted. Then by the Rearrangement Inequality, where we take our indices modulo , with equality exactly when all the , and therefore all the , are equal. Dividing both sides by gives the desired inequality.

### Proof by Cauchy Induction

We first prove that the inequality holds for two variables. Note that are similarly sorted sequences. Then by the Rearrangement Inequality, with equality exactly when .

We next prove that if the inequality holds for variables (with equality when all are equal to zero), than it holds for variables (with equality when all are equal to zero). Indeed, suppose the inequality holds for variables. Let denote the arithmetic means of , , , respectively; let denote their respective geometric means. Then with equality when all the numbers are equal, as desired.

These two results show by induction that the theorem holds for , for all integers . In particular, for every integer , there is an integer such that the theorem holds for variables.

Finally, we show that if and the theorem holds for variables, then it holds for variables with arithmetic mean and geometric mean . Indeed, set for , and let for . Then It follows that , or , with equality exactly when all the are equal to .

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

### Proof by Calculus

We will start the proof by considering the function . We will now find the maximum of this function. We can do this simply using calculus. We need to find the critical points of , we can do that by finding and setting it equal to . Using the linearity of the derivative . We need Note that this is the only critical point of . We can confirm it is the maximum by finding it's second derivative and making sure it is negative. letting x = 1 we get . Since the second derivative , is a maximum. . Now that we have that is a maximum of , we can safely say that or in other words . We will now define a few more things and do some manipulations with them. Let , with this notice that . This fact will come into play later. now we can do the following, let and plug this into , we get Adding all these results together we get Now exponentiating both sides we get This proves the AM-GM inequality.

### Alternate Proof by Induction

Because is a square, . Now, through algebra: From here, we proceed with induction through the second paragraph of the above proof with Cauchy Induction to show the theorem is true.