Proofs of AM-GM
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
Proofs of Unweighted AM-GM
These proofs use the assumption that , for all integers
.
Proof by Cauchy Induction
We use Cauchy Induction, a variant of induction in which one proves AM-GM for , all powers of
, and then that
implies
.
Base Case: The smallest nontrivial case of AM-GM is in two variables. By the properties of perfect squares (or by the Trivial Inequality), with equality if and only if
, or
. Then because
and
are nonnegative, we can perform the following manipulations:
with equality if and only if
, just as before. This completes the proof of the base case.
Powers of Two: We use induction. Suppose that AM-GM is true for variables; we will then prove that the inequality is true for
. Let
be any list of nonnegative reals. Then, because the two lists
and
, have
variables each,
Adding these two inequalities together and dividing by two yields
Then by AM-GM in two variables,
Combining this inequality with the previous one yields unweighted AM-GM in
variables, with one exception — equality.
For equality, note that every inequality mentioned must have equality as well; thus, inequality holds if and only if all the numbers in are the same, all the numbers in
are the same, and
. From here, it is trivial to show that this implies
, which is the equality condition for unweighted AM-GM in
variables.
This completes the induction and implies that unweighted AM-GM holds for all powers of two.
Backward Step: Assume that AM-GM holds for variables. We will then use a substitution to derive AM-GM for
variables. Letting
, we have that
Because we assumed AM-GM in
variables, equality holds if and only if
. However, note that the last equality is implied if all the numbers of
are the same; thus, equality holds if and only if
.
We then simplify the lefthand side. Multiplying both sides of the fraction by and combining like terms, we get that
Thus,
Raising both sides to the
th power yields
From here, we divide by
and take the
root to get that
Every step we took preserves our earlier equality condition, which proves that AM-GM holds for
variables. This completes the backward step, which proves the unweighted AM-GM Inequality.
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 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.
Proof of Weighted AM-GM
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.