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Chapter 6: Problem 250

Problem 250 Let \(a, b\) be real numbers such that \(a \neq b,\) and \(a+b>0\) Prove by induction that $$ 2^{n-1}\left(a^{n}+b^{n}\right) \geqslant(a+b)^{n}, \text { for all } n \geqslant 1 $$

Short Answer

Expert verified
Inductive proof shows the inequality holds for all \(n \geq 1\).

Step by step solution

01

Base Case

First, we test the statement for the base case when \(n = 1\). The inequality becomes \(2^{1-1}(a^1 + b^1) \geq (a+b)^1\), which simplifies to \(a + b \geq a + b\). This statement holds true, thus verifying the base case.
02

Inductive Hypothesis

Assume that the inequality holds for some arbitrary positive integer \(k\), that is, assume \(2^{k-1}(a^k + b^k) \geq (a+b)^k\) is true. This is our inductive hypothesis.
03

Inductive Step

We need to prove that if the inequality is true for \(n = k\), it is also true for \(n = k + 1\). Therefore, we need to show: \ \[ 2^k(a^{k+1} + b^{k+1}) \geq (a+b)^{k+1} \] Using the binomial theorem, express \((a+b)^{k+1}\): \ \[ (a+b)^{k+1} = (a+b)(a+b)^k = (a+b) \sum_{i=0}^{k} \binom{k}{i}a^i b^{k-i} \]
04

Use of the Inductive Hypothesis

By the inductive hypothesis, since \(2^{k-1}(a^k + b^k) \geq (a+b)^k\), we have: \ \[ 2(a^{k+1} + b^{k+1}) \geq a(a+b)^k + b(a+b)^k \] Since multiplying by positive numbers preserves inequality: \ \[ 2^k(a^{k+1} + b^{k+1}) \geq 2(a^{k+1} + b^{k+1}) \geq a(a+b)^k + b(a+b)^k \] Thus, \[2^k(a^{k+1} + b^{k+1}) \geq (a+b)^{k+1}\] holds, closing the induction process.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inequality
An inequality is a mathematical statement that describes the relationship between two values when they are not equal but rather one is greater or less than the other. It uses symbols like <, >, 鈮�, or 鈮� to express this relationship.

In the provided exercise, we need to prove an inequality involving the terms \(2^{n-1}(a^n + b^n)\) and \((a+b)^n\). The goal is to show that the expression on the left side is greater than or equal to the expression on the right for all integers \(n\geq1\).

Inequalities are crucial in mathematics because they help define ranges within which a mathematical expression can fall. They're especially significant in optimization, calculus, and analysis, providing a foundation for establishing limits and bounds.
Real Numbers
Real numbers include all the numbers you can find on the number line. This includes integers, fractions, and irrational numbers. They can be positive, negative, or zero. The significance of real numbers in the problem is in their inclusive nature, which covers any possible value of \(a\) and \(b\) provided they are distinct and their sum is positive.

In our exercise, \(a\) and \(b\) are specified as real numbers. This broad categorization allows the inequality to be examined in general terms rather than being limited to specific kinds of numbers, like just integers or only positive numbers. Such versatility is part of the real number system's power in mathematics, permitting flexible problem-solving across various types of mathematical problems.
Binomial Theorem
The Binomial Theorem is a robust formula used to expand expressions raised to a power, written as \((a + b)^n\). This theorem states that \[(a + b)^n = \sum_{i=0}^{n} \binom{n}{i} a^{n-i} b^i\]where \(\binom{n}{i}\) is a binomial coefficient, calculated as \(\frac{n!}{i!(n-i)!}\).

In the solution, we used the Binomial Theorem to break down \((a+b)^{k+1}\) into simpler components to help prove the inequality at each step. By expressing the binomial expansion, it allows us to manage complex polynomial expressions systematically, focusing on the individual terms, which makes verifying inequalities like in our exercise more tangible.
Inductive Hypothesis
Inductive Hypothesis is a crucial component of the mathematical induction process. When you perform mathematical induction, you make an assumption known as the inductive hypothesis. This is where you assume the statement is true for an arbitrary positive integer \(k\).

In the problem, we assume that \(2^{k-1}(a^k + b^k) \geq (a+b)^k\) holds true for some integer \(k\). This assumption is essential because it serves as the stepping stone for proving the next case, where \(n = k + 1\). After establishing the base case, the inductive hypothesis allows us to build forward, using what we know to show the inequality must also be true in the next step.

This concept is at the heart of mathematical induction, which is a powerful technique for proving propositions or theorems formulated with integers. Understanding and correctly applying the inductive hypothesis is what makes induction one of the most reliable methods in mathematical proof construction.

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Most popular questions from this chapter

Problem 232 The sequence $$ 2,5,13,35, \ldots $$ is defined by its first two terms \(u_{0}=2, u_{1}=5,\) and by the recurrence relation: $$ u_{n+2}=5 u_{n+1}-6 u_{n} $$ (a) Guess a closed formula for the \(n^{\text {th }}\) term \(u_{n}\). (b) Prove by induction that your guess in (a) is correct for all \(n \geqslant 0\).

(a) Prove by induction that $$ \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}} \leqslant 2-\frac{1}{n}, \text { for all } n \geqslant 1 .$$ (b) Prove by induction that $$ \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}<1.68-\frac{1}{n}, \text { for all } n \geqslant 4 $$ The infinite sum $$ \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}+\cdots \text { (for ever) } $$ is a historical classic, and has many instructive stories to tell. Recall that, in Problems \(\mathbf{5 4}, \mathbf{6 2}, \mathbf{6 3}, \mathbf{2 3 6}, \mathbf{2 3 7}, \mathbf{2 3 8}\) you found closed formulae for the sums $$ \begin{array}{c} 1+2+3+\cdots+n \\ 1^{2}+2^{2}+3^{2}+\cdots+n^{2} \\ 1^{3}+2^{3}+3^{3}+\cdots+n^{3} \end{array} $$ and for the sums $$ 1 \times 2+2 \times 3+3 \times 4+\cdots+(n-1) n $$ $$ 1 \times 2 \times 3+2 \times 3 \times 4+3 \times 4 \times 5+\cdots+(n-2)(n-1) n $$ Each of these expressions has a "natural" feel to it, and invites us to believe that there must be an equally natural compact answer representing the sum. In Problem \(\mathbf{2 3 5}\) you took this idea one step further by finding a beautiful closed expression for the sum $$ \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)}=1-\frac{1}{n+1} $$ When we began to consider infinite series, we found the elegant closed formula $$ 1+r+r^{2}+r^{3}+\cdots+r^{n}=\frac{1}{1-r}-\frac{r^{n+1}}{1-r} $$ We then observed that the final term on the RHS could be viewed as an "error term", indicating the amount by which the LHS differs from \(\frac{1}{1-r}\), and noticed that, for any given value of \(r\) between -1 and \(+1,\) this error term "tends towards 0 as the power \(n\) increases". We interpreted this as indicating that one could assign a value to the endless sum $$ 1+r+r^{2}+r^{3}+\cdots(\text { for ever })=\frac{1}{1-r} $$ In the same way, in the elegant closed formula $$ \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)}=1-\frac{1}{n+1} $$ the final term on the RHS indicates the amount by which the finite sum on the LHS differs from \(1 ;\) and since this "error term" tends towards 0 as \(n\) increases, we may assign a value to the endless sum $$ \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots \text { (for ever) }=1 $$ It is therefore natural to ask whether other infinite series, such as $$ \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}+\cdots \text { (for ever) } $$ may also be assigned some natural finite value. And since the series is purely numerical (without any variable parameters, such as the " \(r^{\prime \prime}\) in the geometric series formula), this answer should be a strictly numerical answer. And it should be exact - though all we have managed to prove so far (in Problems 246 and 247 ) is that this numerical answer lies somewhere between \(\frac{17}{12}\) and 1.68 This question arose naturally in the middle of the seventeenth century, when mathematicians were beginning to explore all sorts of infinite series (or "sums that go on for ever"). With a little more work in the spirit of Problems 246 and 247 one could find a much more accurate approximate value. But what is wanted is an exact expression, not an unenlightening decimal approximation. This aspiration has a serious mathematical basis, and is not just some purist preference for elegance. The actual decimal value is very close to $$ 1.649934 \cdots $$ But this conveys no structural information. One is left with no hint as to why the sum has this value. In contrast, the eventual form of the exact expression suggests connections whose significance remains of interest to this day. The greatest minds of the seventeenth and early eighteenth century tried to find an exact value for the infinite sum - and failed. The problem became known as the Basel problem (after Jakob Bernoulli \((1654-1705)\) who popularised the problem in \(1689-\) one of several members of the Bernoulli family who were all associated with the University in Basel). The problem was finally solved in \(1735-\) in truly breathtaking style - by the young Leonhard Euler \((1707-1783)\) (who was at the time also in Basel). The answer $$ \frac{\pi^{2}}{6} $$ illustrates the final sentence of the preceding paragraph in unexpected ways, which we are still trying to understand. In the next problem you are invited to apply similar ideas to an even more important series. Part (a) provides a relatively crude first analysis. Part (b) attacks the same question; but it does so using algebra and induction (rather than the formula for the sum of a geometric series) in a way that is then further refined in part (c).

Let \(f_{1}=2, f_{k+1}=f_{k}\left(f_{k}+1\right)\). Prove by induction that \(f_{k}\) has at least \(k\) distinct prime factors.

(Calkin-Wilf tree) The binary tree in the plane has a distinguished vertex as 'root', and is constructed inductively. The root is joined to two new vertices; and each new vertex is then joined to two further new vertices \(-\) with the construction process continuing for ever (Figure 11 ). Label the vertices of the binary tree with positive fractions as follows: \- the root is given the label \(\frac{1}{1}\) \- whenever we know the label \(\frac{i}{j}\) of a 'parent' vertex, we label its 'left descendant' as \(\frac{i}{i+j},\) and its 'right descendant' \(\frac{i+j}{j}\). (a) Prove that every positive rational \(\frac{r}{s}\) occurs once and only once as a label, and that it occurs in its lowest terms. (b) Prove that the labels are left-right symmetric in the sense that labels in corresponding left and right positions are reciprocals of each other. \(\triangle\)

Problem 259 A tree is a connected graph, or network, consisting of vertices and edges, but with no cycles (or circuits). Prove that a tree with \(n\) vertices has exactly \(n-1\) edges. The next problem concerns spherical polyhedra. A spherical polyhedron is a polyhedral surface in 3-dimensions, which can be inflated to form a sphere (where we assume that the faces and edges can stretch as required). For example, a cube is a spherical polyhedron; but the surface of a picture frame is not. A spherical polyhedron has \- faces (flat 2-dimensional polygons, which can be stretched to take the form of a disc), \- edges (1-dimensional line segments, where exactly two faces meet), and \- vertices (0-dimensional points, where several faces and edges meet, in such a way that they form a single cycle around the vertex). Each face must clearly have at least 3 edges; and there must be at least three edges and three faces meeting at each vertex. If a spherical polyhedron has \(V\) vertices, \(E\) edges, and \(F\) faces, then the numbers \(V, E, F\) satisfy Euler's formula $$ V-E+F=2 $$ For example, a cube has \(V=8\) vertices, \(E=12\) edges, and \(F=6\) faces, and \(8-12+6=2\)
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