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Chapter 13: Problem 30

Prove that if \(x_{1} y_{1}=1\) and \(x_{2} y_{2}=1\) then their average \(x=\frac{1}{2}\left(x_{1}+x_{2}\right), y=\frac{1}{2}\left(y_{1}+y_{2}\right)\) has \(x y \geqslant 1 .\) The function \(f=x y\) has convex level curves (hyperbolas).

Short Answer

Expert verified
The product of the averages satisfies \(xy \geq 1\) due to the convexity of \(f = xy\).

Step by step solution

01

Understand the Problem

We need to show that the average values of two pairs \((x_1, y_1)\) and \((x_2, y_2)\) that satisfy \(x_1 y_1 = 1\) and \(x_2 y_2 = 1\) will have their product \(xy \geqslant 1\). We also know that the level curves of the function \(f = xy\) are hyperbolas.
02

Expand and Simplify

Start with the expressions for the averages: \(x = \frac{1}{2}(x_1 + x_2)\) and \(y = \frac{1}{2}(y_1 + y_2)\). Now, calculate the product: \(xy = \left(\frac{1}{2}(x_1 + x_2)\right)\left(\frac{1}{2}(y_1 + y_2)\right)\). Simplify this to: \(xy = \frac{1}{4}(x_1y_1 + x_1y_2 + x_2y_1 + x_2y_2)\).
03

Substitute Known Values

Substitute the given conditions \(x_1 y_1 = 1\) and \(x_2 y_2 = 1\) into the expression:\[xy = \frac{1}{4}(1 + x_1y_2 + x_2y_1 + 1) = \frac{1}{4}(2 + x_1y_2 + x_2y_1)\].
04

Apply Cauchy-Schwarz Inequality

Use the Cauchy-Schwarz inequality in the form \((a^2 + b^2)(c^2 + d^2) \geq (ac + bd)^2\). Applying it here for \((a, b) = (x_1, x_2)\) and \((c, d) = (y_2, y_1)\), we get \((x_1^2 + x_2^2)(y_2^2 + y_1^2) \geq (x_1y_2 + x_2y_1)^2\). Recognizing that each product on the left equals 2 since \((x_1 y_1 = 1)\) and \((x_2 y_2 = 1)\), we conclude that \(2 \cdot 2 \geq (x_1y_2 + x_2y_1)^2\).
05

Complete the Proof

Taking square roots gives \(4 \geq x_1y_2 + x_2y_1\), hence \(\frac{1}{4}(2 + x_1y_2 + x_2y_1) \geq \frac{1}{4}(2 + 2) = 1\). Thus, \(xy \geq 1\) is true, as we needed to show.

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Key Concepts

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

Convex Level Curves
A level curve is a set of points where a function takes on a certain constant value. In mathematical terms, it's like drawing a contour on a map where all points along the contour share the same elevation. For the function \(f = xy\), the level curves are defined by the equation \(xy = c\), where \(c\) is a constant.
These curves are shaped like hyperbolas, which is a type of smooth curve lying in a plane. They appear in pairs and are typically symmetric.
  • Convex level curves, like those of \(xy = c\), have a distinctive property: they bend outward, kind of like a bowl turned upside down.
  • Convexity means that if you pick any two points on a curve, the line segment that joins them will lie entirely above or on the curve.
Understanding these curves helps us visualize how the function behaves, which can be critical when proving problems involving inequalities or averages, such as the one involving two pairs of numbers where their products equal one.
Hyperbolas
Hyperbolas are fascinating shapes that arise in the world of conic sections. They are defined as the set of points for which the absolute difference of the distances to two fixed points (foci) is constant.
For our function \(f = xy\), the equations \(xy = 1\) and \(xy = 2\) are examples of hyperbolas.
  • Each hyperbola consists of two disconnected curves known as branches.
  • In the coordinate plane, a hyperbola can be seen opening either horizontally or vertically.
By understanding hyperbolas, we can better approach problems that require visualizing how some parameters of the function change while others stay constant. This is crucial when working with level curves that form part of the mathematical proof in our exercise.
Mathematical Proof
A mathematical proof is a series of logical steps that show why a statement is true. The process involves starting from what we know and using rules and theorems to arrive at what we want to prove.
In our exercise, the proof uses the Cauchy-Schwarz Inequality:
  • This powerful inequality is essential in many areas of mathematics and states that for any sequences of real numbers \((a_i)\) and \((b_i)\), we have: \((\sum a_i^2)(\sum b_i^2) \geq (\sum a_i b_i)^2\).
  • It helps compare sums of powers, providing a way to bound expressions from above.
Applying the Cauchy-Schwarz Inequality to \(x_1, x_2\) and \(y_1, y_2\) in our problem, we found the limits of their sums' products. Proving \(xy \geq 1\) involved recognizing these bounds and showing the steps that ensure all conditions were met. This validates the average's relationship to the initial pairs \(x_1 y_1 = 1\) and \(x_2 y_2 = 1\), demonstrating the elegance and necessity of structured proof.

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Most popular questions from this chapter

True or false, when \(f(x, y)\) is any smooth function: (a) There is a direction \(\mathbf{u}\) at \(P\) in which \(D_{u} f=0\) (b) There is a direction \(\mathbf{u}\) in which \(D_{u} f=\operatorname{grad} f\) (c) There is a direction \(\mathbf{u}\) in which \(D_{\mathbf{q}} f=1\). (d) The gradient of \(f(x) g(x)\) equals \(g\) grad \(f+f \operatorname{grad} g\).

Compute grad \(f,\) then \(D_{\mathrm{a}} f=(\operatorname{grad} f) \cdot \mathrm{u},\) then \(D_{\mathrm{u}} f\) at \(P\). \(f(x, y)=3 x+4 y+7 \quad \mathbf{u}=(3 / 5,4 / 5) \quad P=(0, \pi / 2)\)

Apply Newton's method to a linear problem: \(g=\) \(x+2 y-5=0, h=3 x-3=0 .\) From any starting point show that \(\left(x_{1}, y_{1}\right)\) is the exact solution (convergence in one step).

Find the plane \(z=a x+b y+c\) that best fits the points \((x, y, z)=(0,0, t),(1,0,0),(1,1,2),(0,1,2)\). The answer \(a, b, c\) minimizes the sum of \((z-a x-b y-c)^{2}\) at the four points.

Compute grad \(f,\) then \(D_{\mathrm{a}} f=(\operatorname{grad} f) \cdot \mathrm{u},\) then \(D_{\mathrm{u}} f\) at \(P\). \(f(x, y)=y^{10} \quad \mathbf{u}=(0,-1) \quad P=(1,-1)\)
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