Keo's Notes

Inequalities

7 min read

Algebra

Growth Rate of Functions

Polynomials

Exponential Functions

Weighted F-Mean (Mean in an Arbitrary Function)

Given any function which is both Continuous and Injective (So that is well defined), The f-mean of with weights is defined as

Since is both Continuous and Injective it follows that is strictly monotonic so it follows that .

Weighted Jensen’s Inequality

Let be an interval on the reals be Convex on , with and real weights (In many scenarios the weights will be normalised such that ), Then

If is concave then the inequality is flipped. The equality holds if or is affine on the on a domain containing . Proof

  1. W.L.O.G let .
  2. Trivially for so and the (in)equality holds.
  3. The definition of convexity means . This is the case.
  4. Assume that Jensen’s holds for , then for with weights (arbitrary, not necessarily the same as the case) such that . If then all other terms are zero so the inequality trivially holds. If it is equivalent to the induction hypothesis so assume
  5. Let
  6. So we can rewrite
  8. By convexity (same as case),
  9. By our inductive hypothesis
  10. So we get that
  11. The exact same argument follows for concave functions but the inequality of the case flips .

Intuition: The way this proof works is that we view the last point separately and the rest as one convex combination. We then renormalise with to insure the weights sum to (ensuring that the one point a convex combination) so we can apply the two-point Jensen that states that and we’ve just used . And then invoke our induction hypothesis on . ( acts as a shortcut for all input as it forms a new point which lies in the interval as ) Online Proof

Weighted Power Mean

The weighted power mean is the case of and of the f-mean. Proof

Given with weights where and , The weighted power mean is

The weights serve to emphasises a certain element so if is more important then will be greater. In the case where , we get an unweighted/regular version. Some special case for are: . is valid for .

Weighted Power Mean Inequality

The equality holds if and only if . For the cases of we get the inequality . Again the equality only holding when all elements are equal.

Weighted Geometric Mean

The weighted geometric mean is the case of the f-mean where and Proof

  2. Using logarithm laws we get that
  3. So
  4. W.L.O.G we can assume that to get a cleaner version and obtain

It is also the of the power mean Proof

  4. indeterminate form we can use L’Hopitals Rule
  6. Therefore

Muirhead’s Inequality

Suppose we have two sequences and such that and , we say that also denoted as . Muirhead’s Inequality states that

INSERT PROOF with Karamta’s Inequality

Holders Inequality

For sequences and weights , Holders states that

With the equality holding if and only if the sequences are proportional to each other INSERT PROOF

Variants

Case

Using Holders with two sequences and weights for Let and and , Then

With the equality only holding when the sequences are proportional to each other. Proof

  1. Plugging in conditions to Holders we get
  2. Rearranging and raising both sides to the power of gets as desired .

Alternate Proof

  1. Since the inequality is homogeneous we can scale both and such that
  2. Then by Weighted AM-GM

More concretely the scaling that is used is and . So we yield that and So both sides scale by a positive factor of so we can divide through and now we have

Classical Conjugate-Exponent Form

Using the case and making transformations and renaming exponents so that (This adds the condition ) we get

Cauchy-Schwarz

Using the classical conjugate-exponent form and setting then squaring the whole inequality (Also equivalent to the full holders inequality with then transformations )

Note that now we introduced the squares we have expanded the domain so now we get that

Titu’s Lemma

Titu’s Lemma states that for and

Note this is just Cauchy-Schwarz applied to the sequences and hence the extra restrictions on . This is a very useful version for eliminating fractions.

Chebyshev’s Sum Inequality

Let and such that they are both monotonic in the same direction, Then

With equality only holding when at least on of the sequences is constant or they are proportional meaning . If they are monotonic but in opposite directions then the reverse inequality holds with the same conditions for equality. INSERT PROOF

Schur’s Inequality

Let , Then Schur’s Inequality States

With equality if and only if or The case is most common and is

INSERT PROOF

Tangent Line Trick

Graph View

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