Prove The Following Inequalities For Real Numbers X, Y, And Z A) X^2 + Y^2 + Z^2 + 3/2 >= Sqrt(2)*(x + Y + Z); B) X^3 + Y^3 + Z^3 >= 3xyz, Where X, Y, Z Are In (0, +infinity); C) (|x+y|)/(1+|x+y|) <= |x|/(1+|x|) + |y|/(1+|y|).

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This article delves into the fascinating world of inequalities, focusing on three specific problems involving real numbers. Inequalities play a crucial role in various branches of mathematics, including calculus, analysis, and optimization. Mastering the techniques to prove inequalities is essential for any aspiring mathematician or problem-solver. In this exploration, we will rigorously demonstrate the validity of each inequality, providing detailed explanations and insights along the way. We aim to enhance your understanding of inequalities and equip you with the necessary tools to tackle similar challenges. Our focus will be on providing clear, concise, and comprehensive proofs that are accessible to a wide audience. We will employ a combination of algebraic manipulations, well-known inequality theorems, and logical reasoning to establish the results. Through this journey, we hope to not only solve these specific problems but also to foster a deeper appreciation for the elegance and power of mathematical inequalities.

a) Proving x² + y² + z² + 3/2 ≥ √2(x + y + z)

To demonstrate the inequality x² + y² + z² + 3/2 ≥ √2(x + y + z), where x, y, and z are real numbers, we will employ a strategic approach that combines algebraic manipulation and the application of fundamental inequality principles. Our goal is to transform the given inequality into a more manageable form, ultimately revealing its inherent truth. We begin by rearranging the terms to bring all expressions to one side, facilitating a clearer analysis. This rearrangement sets the stage for applying a technique that involves completing the square, a powerful method for dealing with quadratic expressions. By completing the square, we aim to rewrite the left-hand side of the inequality in a way that highlights its non-negativity, a crucial step in proving the desired result. The process of completing the square will lead us to a form where we can easily observe that the expression is always greater than or equal to zero, thus establishing the validity of the original inequality. This method not only proves the inequality but also provides a deeper understanding of the relationship between the variables and constants involved. The elegance of this approach lies in its ability to transform a seemingly complex inequality into a straightforward assertion that can be readily verified. Let's embark on this journey, step by step, to unravel the beauty of this mathematical proof.

Proof

We aim to prove that x² + y² + z² + 3/2 ≥ √2(x + y + z) for all real numbers x, y, and z. Let's start by rearranging the inequality:

x² + y² + z² + 3/2 - √2(x + y + z) ≥ 0

Now, let's rewrite this as:

(x² - √2x) + (y² - √2y) + (z² - √2z) + 3/2 ≥ 0

We will complete the square for each quadratic term. Recall that to complete the square for a quadratic expression of the form ax² + bx, we add and subtract (b/2a)².

For x² - √2x, we add and subtract (√2/2)² = 1/2. For y² - √2y, we add and subtract (√2/2)² = 1/2. For z² - √2z, we add and subtract (√2/2)² = 1/2.

So, we have:

(x² - √2x + 1/2) + (y² - √2y + 1/2) + (z² - √2z + 1/2) + 3/2 - 1/2 - 1/2 - 1/2 ≥ 0

This simplifies to:

(x - √2/2)² + (y - √2/2)² + (z - √2/2)² ≥ 0

Since squares of real numbers are non-negative, the sum of squares is also non-negative. Thus, the inequality holds true for all real numbers x, y, and z.

b) Proving x³ + y³ + z³ ≥ 3xyz for x, y, z in (0, +∞)

To establish the inequality x³ + y³ + z³ ≥ 3xyz, where x, y, and z are positive real numbers, we will invoke the powerful AM-GM inequality (Arithmetic Mean - Geometric Mean). This inequality provides a fundamental relationship between the arithmetic mean and the geometric mean of a set of non-negative numbers. It states that the arithmetic mean is always greater than or equal to the geometric mean. By skillfully applying the AM-GM inequality to the terms x³, y³, and z³, we can directly demonstrate the desired result. The beauty of this approach lies in its simplicity and elegance, showcasing how a fundamental principle can be used to solve a seemingly complex problem. The AM-GM inequality is a cornerstone of mathematical analysis, and its application here highlights its versatility and power. This method not only proves the inequality but also provides a deeper understanding of the relationship between the cubes of positive numbers and their product. Let us delve into the application of this inequality, revealing the straightforward path to proving the given assertion. The AM-GM inequality is not just a tool for solving problems; it is a gateway to understanding the intricate relationships that exist within the world of numbers and quantities.

Proof

We will use the AM-GM inequality, which states that for non-negative real numbers a₁, a₂, ..., aₙ, the following holds:

(a₁ + a₂ + ... + aₙ)/n ≥ ⁿ√(a₁a₂...aₙ)

Applying the AM-GM inequality to x³, y³, and z³ (which are positive since x, y, z are in (0, +∞)), we have:

(x³ + y³ + z³)/3 ≥ ³√(x³y³z³)

This simplifies to:

(x³ + y³ + z³)/3 ≥ xyz

Multiplying both sides by 3, we get:

x³ + y³ + z³ ≥ 3xyz

Thus, the inequality holds true for all positive real numbers x, y, and z.

c) Proving (|x+y|)/(1+|x+y|) ≤ |x|/(1+|x|) + |y|/(1+|y|)

To establish the inequality (|x+y|)/(1+|x+y|) ≤ |x|/(1+|x|) + |y|/(1+|y|), we will leverage a crucial property of the absolute value function and a clever algebraic manipulation. This inequality showcases the interplay between absolute values and fractions, highlighting how a careful analysis can lead to a concise and elegant proof. Our approach will begin by employing the triangle inequality, a fundamental principle that states |x + y| ≤ |x| + |y|. This inequality serves as the cornerstone of our argument, providing a critical link between the left-hand side and the right-hand side of the given inequality. Subsequently, we will introduce a function f(t) = t/(1+t), where t is a non-negative real number. By demonstrating that this function is subadditive, meaning f(a+b) ≤ f(a) + f(b), we can effectively bridge the gap between the expressions involving absolute values. This subadditivity property is key to proving the desired result, as it allows us to decompose the complex inequality into simpler, more manageable components. The beauty of this method lies in its strategic use of established principles and functions to elegantly solve the problem. This proof not only validates the inequality but also deepens our understanding of the properties of absolute values and their interactions with other mathematical constructs. Let's embark on this journey, dissecting the problem step by step and unveiling the logical flow that leads to the solution.

Proof

Let's consider the function f(t) = t/(1+t) for t ≥ 0. We want to show that f(t) is subadditive, i.e., f(a+b) ≤ f(a) + f(b) for non-negative a and b.

First, note that f(t) can be rewritten as f(t) = 1 - 1/(1+t). This form makes it easier to analyze the behavior of the function.

Now, let's consider the inequality we want to prove:

(|x+y|)/(1+|x+y|) ≤ |x|/(1+|x|) + |y|/(1+|y|)

Using the triangle inequality, we know that |x + y| ≤ |x| + |y|.

Let a = |x| and b = |y|. We want to show that f(|x+y|) ≤ f(|x|) + f(|y|), which is equivalent to showing f(|x+y|) ≤ f(a) + f(b).

Let's prove that f(t) is subadditive. We need to show that:

(a+b)/(1+a+b) ≤ a/(1+a) + b/(1+b)

Multiplying both sides by (1+a+b)(1+a)(1+b), we get:

(a+b)(1+a)(1+b) ≤ a(1+a+b)(1+b) + b(1+a+b)(1+a)

Expanding the terms:

(a+b)(1 + a + b + ab) ≤ a(1 + b + a + ab + b + b²) + b(1 + a + b + ab + a + a²)

a + a² + ab + a²b + b + ab + b² + ab² ≤ a + ab + a² + a²b + ab + ab² + b + ab + b² + ab² + ab + a²b

a + a² + 2ab + a²b + b + b² + ab² ≤ a + a² + 2ab + a²b + b + b² + ab² + ab

Simplifying:

a²b + ab² ≤ ab + a²b + ab²

0 <= ab

Now, we return the original inequality by applying the function f:

f(|x+y|) = (|x+y|)/(1+|x+y|)

Since |x+y| ≤ |x| + |y|, we have:

f(|x+y|) ≤ f(|x| + |y|)

And we know that f(|x| + |y|) ≤ f(|x|) + f(|y|), so:

f(|x+y|) ≤ f(|x|) + f(|y|)

Therefore:

(|x+y|)/(1+|x+y|) ≤ |x|/(1+|x|) + |y|/(1+|y|)

Thus, the inequality holds true for all real numbers x and y.

In this article, we have successfully demonstrated three distinct inequalities involving real numbers. Each proof employed a unique approach, showcasing the diverse techniques available for tackling such problems. The first inequality, x² + y² + z² + 3/2 ≥ √2(x + y + z), was proven by completing the square, a method that transforms quadratic expressions into a form that readily reveals their non-negativity. The second inequality, x³ + y³ + z³ ≥ 3xyz for positive real numbers x, y, and z, was elegantly established using the AM-GM inequality, highlighting the power of this fundamental principle in mathematical analysis. Finally, the third inequality, (|x+y|)/(1+|x+y|) ≤ |x|/(1+|x|) + |y|/(1+|y|), was proven by leveraging the triangle inequality and demonstrating the subadditivity of a carefully chosen function. These proofs not only validate the inequalities but also provide valuable insights into the underlying mathematical concepts and techniques. By mastering these approaches, readers can enhance their problem-solving skills and deepen their appreciation for the beauty and rigor of mathematical reasoning. The exploration of inequalities is a continuous journey, and the principles and methods discussed here serve as a solid foundation for tackling more complex and challenging problems in the future. We hope this article has been both informative and inspiring, encouraging further exploration of the fascinating world of mathematics.