Proving Inequality For Real Numbers A Trigonometric Approach

Hey guys! Let's dive deep into the fascinating world of real numbers and tackle a super interesting inequality problem. We're going to show that for any real numbers x and y, the expression $\frac{2(x-y)(1+xy)}{(1+x^2)(1+y^2)}$ always falls within the cozy range of -1 to 1. Sounds like a challenge? Absolutely! But don't worry, we'll break it down step by step and make it crystal clear.

Before we jump into the nitty-gritty details, let's take a moment to appreciate the beauty of inequalities. They're not just about comparing numbers; they're the backbone of optimization problems, the foundation of many mathematical proofs, and even pop up in real-world applications like economics and engineering. So, understanding inequalities is a seriously valuable skill to have in your mathematical toolkit.

Now, let's set the stage. We're given two real numbers, x and y. Remember, real numbers are all the numbers you can think of on a number line – positive, negative, fractions, decimals, even those crazy irrational numbers like pi! Our mission is to prove that no matter what values x and y take, the expression $\frac{2(x-y)(1+xy)}{(1+x^2)(1+y^2)}$ will always be between -1 and 1, inclusive. That means it can be -1, it can be 1, or it can be any number in between. Intrigued? Let's get started!

The Trigonometric Transformation The Key to Unlocking the Inequality

Here's a clever trick that will make our lives much easier: we're going to use a trigonometric substitution. This might sound intimidating, but trust me, it's a powerful technique that often simplifies complex expressions. The idea is to replace x and y with trigonometric functions. Specifically, we'll let $x = \tan(a)$ and $y = \tan(b)$, where a and b are angles in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$. Why this range? Because the tangent function covers all real numbers within this interval, ensuring we don't miss any possible values of x and y.

Think about it: the tangent function is periodic, meaning it repeats its values over and over again. But within the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, the tangent function has a one-to-one correspondence with the real numbers. This means that for every real number x, there's exactly one angle a in this interval such that $x = \tan(a)$, and vice versa. Similarly, for every real number y, there's a unique angle b in the same interval such that $y = \tan(b)$.

Now, why did we choose the tangent function? Well, it turns out that the tangent function has some beautiful trigonometric identities that will help us simplify our expression. Remember those trig identities from high school? They're about to make a comeback! Specifically, we'll be using the tangent addition and subtraction formulas. But before we get there, let's see what happens when we substitute $x = \tan(a)$ and $y = \tan(b)$ into our expression.

Our expression becomes $\frac{2(\tan(a) - \tan(b))(1 + \tan(a)\tan(b))}{(1 + \tan^2(a))(1 + \tan^2(b))}$. Don't panic! It might look even more complicated now, but we're about to work our magic. Remember the Pythagorean identity: $1 + \tan^2(\theta) = \sec^2(\theta)$? We can use this to simplify the denominator. Our expression now transforms into $\frac{2(\tan(a) - \tan(b))(1 + \tan(a)\tan(b))}{\sec^2(a)\sec^2(b)}$. See? We're already making progress!

Unveiling the Tangent Subtraction Formula A Crucial Step

Now comes the moment we've been waiting for: the tangent subtraction formula. This formula states that $\tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)}$. Take a close look at our numerator. It has a term $(\tan(a) - \tan(b))$ and a term $(1 + \tan(a)\tan(b))$. These are exactly the ingredients we need to use the tangent subtraction formula! We can rewrite the numerator as $2\tan(a - b)(1 + \tan(a)\tan(b))$.

But wait, there's more! Remember that $\sec(\theta) = \frac{1}{\cos(\theta)}$? So, $\sec^2(\theta) = \frac{1}{\cos^2(\theta)}$. Using this, we can rewrite the denominator as $\frac{1}{\cos^2(a)\cos^2(b)}$. Now, our expression looks like $\frac{2\tan(a - b)(1 + \tan(a)\tan(b))}{\frac{1}{\cos^2(a)\cos^2(b)}}$. This might still seem a bit messy, but we're getting closer to a simplified form.

To clean things up, we can multiply the numerator and denominator by $\cos^2(a)\cos^2(b)$. This gives us $2\tan(a - b)(1 + \tan(a)\tan(b))\cos^2(a)\cos^2(b)$. Now, let's think about how we can further simplify this. Remember that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$? We can rewrite $\tan(a - b)$ as $\frac{\sin(a - b)}{\cos(a - b)}$. Substituting this in, our expression becomes $2\frac{\sin(a - b)}{\cos(a - b)}(1 + \tan(a)\tan(b))\cos^2(a)\cos^2(b)$.

The Grand Finale Simplifying to the Sine of a Double Angle

Okay, guys, we're in the home stretch! Let's focus on the term $(1 + \tan(a)\tan(b))\cos^2(a)\cos^2(b)$. We can rewrite this as $\cos^2(a)\cos^2(b) + \tan(a)\tan(b)\cos^2(a)\cos^2(b)$. Now, let's substitute $\tan(a) = \frac{\sin(a)}{\cos(a)}$ and $\tan(b) = \frac{\sin(b)}{\cos(b)}$. We get $\cos^2(a)\cos^2(b) + \frac{\sin(a)}{\cos(a)}\frac{\sin(b)}{\cos(b)}\cos^2(a)\cos^2(b)$.

Notice that we can cancel out some cosines! This simplifies to $\cos^2(a)\cos^2(b) + \sin(a)\sin(b)\cos(a)\cos(b)$. Now, we can factor out a $\cos(a)\cos(b)$ to get $\cos(a)\cos(b)(\cos(a)\cos(b) + \sin(a)\sin(b))$. Hey, that looks familiar! Remember the cosine subtraction formula: $\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)$? We can substitute this in, and our term becomes $\cos(a)\cos(b)\cos(a - b)$.

Now, let's put everything back together. Our entire expression is now $2\frac{\sin(a - b)}{\cos(a - b)}\cos(a)\cos(b)\cos(a - b)$. Look at that! We can cancel out the $\cos(a - b)$ terms, leaving us with $2\sin(a - b)\cos(a)\cos(b)$.

But we're not quite done yet. Remember the sine addition formula: $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$? We can use this to simplify further. Our expression now becomes $\sin(2(a - b))$. Wow! That's a lot simpler than where we started.

Now, here's the final piece of the puzzle. We know that the sine function always takes values between -1 and 1. That is, for any angle $\theta$, $-1 \leq \sin(\theta) \leq 1$. Therefore, $-1 \leq \sin(2(a - b)) \leq 1$. And that's exactly what we wanted to show! We've successfully proven that $\frac{2(x-y)(1+xy)}{(1+x^2)(1+y^2)}$ is always between -1 and 1.

Conclusion A Triumph of Trigonometric Transformations

So, there you have it, guys! We've conquered this inequality using a clever trigonometric substitution and a bunch of trigonometric identities. This problem beautifully illustrates the power of these techniques in simplifying complex expressions and solving seemingly difficult problems. The key takeaway here is to not be afraid to try different approaches and to leverage the tools you have in your mathematical arsenal. Trigonometric substitutions can often be a game-changer, especially when dealing with expressions involving squares and sums.

Remember, math is like a puzzle. Sometimes, you need to try different pieces and rotate them in different ways before you find the perfect fit. And sometimes, the solution lies in a clever trick or a unexpected transformation. Keep exploring, keep experimenting, and keep having fun with math! You'll be amazed at what you can achieve.

I hope this explanation was clear and helpful. If you have any questions or want to explore more inequality problems, feel free to ask! Keep up the great work, and I'll see you in the next mathematical adventure!