Inequalities
[41/85%]Prove that for any finite sequence of real numbers
there exists an index such that
(for the first sum is taken to be zero, for — the second);
the inequality
holds (the sum is over all possible sign assignments ).
Let and be finite nonnegative sequences, with the sequence nonincreasing; . Prove that if the indices do not exceed and satisfy the inequality , then
Let be a finite sequence of positive numbers and . Prove that
;
.
Let be a finite sequence of numbers, with for all . Prove the inequality
Let be a finite sequence of nonnegative numbers, with . Prove the inequalities
the right-hand inequalities are strict if ; the left-hand inequalities are strict if at least two of the numbers are positive.
Let be a finite sequence of real numbers, with for . Prove that
if , then
equality is possible only in the cases where or ;
if , then
equality is possible only in the case where .
Let and be finite sequences of nonnegative numbers. Prove that
if , then ;
(Minkowski's inequality).
Let be a finite sequence of complex numbers, and . Prove that and .
Let . Prove that for all the inequalities
hold.
For which are these inequalities sharp in order as increases without bound?
Prove that
for ;
for .
From these inequalities it follows that , where . More precise representations of the factorial are considered in problems II.2.9 and II.2.10.
is the sequence of prime numbers, numbered in increasing order (). Let . Prove that
;
for .
is the sequence of prime numbers, numbered in increasing order (). Prove that for the following inequalities hold:
;
.
is the sequence of prime numbers, numbered in increasing order (). Let be the number of primes not exceeding . Verify that
Deduce from this that for all the following inequalities hold:
;
;
;
.
is the sequence of prime numbers, numbered in increasing order (). Let be the least common multiple of the numbers , . Verify that
;
.
Using these relations, prove that the estimates for and obtained in problems 2.13 c), d) are sharp in order:
is the sequence of prime numbers, numbered in increasing order (). Prove that
For finite sequences of real numbers and , denote by and (respectively and ) the nondecreasing (respectively nonincreasing) rearrangements of these sequences. Prove the inequalities
;
.
In solving the following problems the Abel transformation (the discrete analogue of the integration-by-parts formula) may prove useful:
where . This equality is especially convenient to use in those cases where the sequence is monotone.
Let be a finite sequence of real numbers. Prove that for any finite nonnegative nonincreasing sequence the following inequalities hold:
;
;
,
where and . Here the factor cannot be replaced by a smaller one, and the factor cannot be replaced by a larger one.
Let be a finite sequence of real numbers. Prove that
where . In which cases do these inequalities become equalities?
Let be a nondecreasing convex (or nonincreasing concave) sequence of real numbers. Prove the inequalities
If the sequence is nonincreasing and convex (or nondecreasing and concave), then the inequality signs should be reversed.
The Chebyshev polynomial is the polynomial defined on the interval by the equality
(the graphs of for are shown in Fig. 4, and for in Fig. 5).
Fig. 4
Fig. 5
Prove that
is a polynomial of degree ;
for the leading coefficient of equals ;
the Chebyshev polynomials satisfy the recurrence relation ;
the Chebyshev polynomials satisfy the differential equation ;
for the Chebyshev polynomials satisfy the equality
the fraction is a generating function for the Chebyshev polynomials, i.e.
More detailed information on Chebyshev polynomials can be found in the book [ Pa].
This problem uses the Chebyshev polynomials (see problem 2.20). Prove the extremal property of Chebyshev polynomials: if is an algebraic polynomial of degree (with real or complex coefficients) and for all , then for all and . The inequality is strict if , where .
In particular, for we obtain outside the interval , and the inequality for implies that the leading coefficient of the polynomial is, in absolute value, at most , with equality possible only for , where .
This problem uses the Chebyshev polynomials (see problem 2.20). Let be a trigonometric polynomial of order with real coefficients and . Set . Prove Bernstein's inequality:
Deduce from this Bernstein's inequality for algebraic polynomials:
where is an algebraic polynomial of degree with real coefficients, .
In which cases do these inequalities turn into equalities at at least one point?
Let . Prove the inequalities
for ;
for .
Let and . Prove that the function is nonnegative on the interval and
The next problem considers relations that supplement the classical inequalities
Prove the inequalities
for ;
for ;
for ;
for ;
for .
Does the inequality hold on for some fixed ?
Prove that on the interval the inequalities hold
;
.
Prove the inequalities
for ;
for ;
for .
What happens for ?
for .
What happens for ?
Prove the inequalities
Prove that for any and the inequalities hold
;
;
.
Prove that for .
Prove that for any the following inequalities hold
;
;
;
.
In the case , the signs of the inequalities in (b)-(d) are reversed.
Let , where the function is nondecreasing and on . Prove the inequality
The integral inequalities considered here have obvious summation analogues; their proofs can be obtained either by passing to the limit in the summation inequality, or by modifying the proofs used in the discrete case. Let the function be integrable on the interval and . Then the following inequalities hold:
The integral inequalities considered here have obvious summation analogues; their proofs can be obtained either by passing to the limit in the summation inequality, or by modifying the proofs used in the discrete case. By means of passage to the limit these inequalities carry over to the case of improper integrals over the interval . Let the functions and be integrable on the interval , with . For set . Then the following inequalities hold:
;
;
The integral inequality considered here has an obvious summation analogue; its proof can be obtained either by passing to the limit in the summation inequality, or by modifying the proof used in the discrete case. Let the functions and be integrable on the interval , with and . Set . Then
The integral inequality considered here has an obvious summation analogue; its proof can be obtained either by passing to the limit in the summation inequality, or by modifying the proof used in the discrete case. Let . Then
Equality holds only for linear functions.
The integral inequalities considered here have obvious summation analogues; their proofs can be obtained either by passing to the limit in the summation inequality, or by modifying the proofs used in the discrete case. If a continuously differentiable function is nondecreasing and convex (or nonincreasing and concave) on the interval , then
For a nondecreasing concave (or nonincreasing convex) function the signs of the inequalities are reversed. Equality is possible only for a linear function.
The integral inequalities considered here have obvious summation analogues; their proofs can be obtained either by passing to the limit in the summation inequality, or by modifying the proofs used in the discrete case. Let the functions and be positive and integrable on the interval . Then the following inequalities hold:
(an analogue of the Cauchy inequality between the arithmetic mean and the geometric mean);
(an analogue of the Minkowski inequality — see problem 2.7 (b)).
The integral inequality considered here has an obvious summation analogue; its proof can be obtained either by passing to the limit in the summation inequality, or by modifying the proof used in the discrete case. Let the functions and be monotonic on the interval , and let the function be nonnegative and integrable on it. If the monotonicity of the functions is of the same type, then the following inequality holds:
while if the monotonicity is of different types, then the sign of the inequality is reversed.
-
Let , with . Prove the inequality
by representing the difference of these integrals as an integral of a nonnegative function.
-
Prove that the inequality remains valid if, instead of the condition , we assume that and .
-
Let and . Prove that
For which functions do these inequalities become equalities?
Let a smooth closed curve of length bound a plane figure of area . Using the natural parametrization of the curve and Wirtinger's inequality, prove the isoperimetric inequality: . Use it to determine the maximal area of a figure that can be bounded by a curve of length , lying in a half-plane, with endpoints on the boundary of the half-plane — "Dido's problem".
According to legend, the ship of Dido, fleeing the persecution of her brother, the Tyrian king Pygmalion, landed on the African shore. The local king Iarbas, in response to the fugitives' request for land to settle on, allowed them to occupy a plot on the seashore of the size of an ox hide. Dido ordered the hide to be cut into thin strips and used them to enclose a plot of land on which she founded Carthage.