Inequality Puzzle: Finding Real Number Triplets

Aug 12, 2025 by RICHARD 48 views

Unraveling the Mystery: How Many Triplets Satisfy the Inequality?

Hey guys! Ever stumbled upon a math problem that just makes you scratch your head? Well, I recently encountered one that had me doing just that. It's all about finding real number triplets that satisfy a certain inequality. Sounds intimidating, right? But don't worry, we're going to break it down together, step by step, in a way that's hopefully not too painful. Think of it as a mathematical puzzle we're solving as a team! So, grab your favorite beverage, and let's dive into this intriguing problem.

The Challenge: Decoding the Inequality

So, what's this mathematical beast we're trying to tame? Here it is:

2x√(y-1) + 2y√(z-1) + 2z√(x-1) ≥ xy + yz + zx

Our mission, should we choose to accept it, is to figure out how many triplets (x, y, z) of real numbers actually make this inequality true. At first glance, it looks like a jumble of variables and square roots. Where do we even start, right? But fear not! We're going to dissect this thing like a frog in biology class (except hopefully less messy and more insightful!). We will explore effective techniques for solving inequalities.

Initial Thoughts: Setting the Stage

The very first thing that probably jumps out at you (or at least it should!) is those square roots. We've got √(y-1), √(z-1), and √(x-1). Now, remember your basic math rules: you can't take the square root of a negative number (at least not in the realm of real numbers). So, that immediately tells us something crucial: x, y, and z must all be greater than or equal to 1. This is our foundation, the ground rules of the game. If any of them are less than 1, the whole thing falls apart. This initial observation narrows down our search considerably. We're not dealing with the entire vast universe of real numbers, just the ones hanging out at 1 or above. This is a crucial first step in problem-solving: identifying constraints.

Digging Deeper: The AM-GM Inequality to the Rescue

Now that we know x, y, and z are at least 1, we can start thinking about how to tackle the inequality itself. This is where some clever mathematical tools come into play. One of the most powerful tools in the inequality-solving toolbox is the AM-GM inequality. This might sound like some kind of secret code, but it's actually quite straightforward. AM-GM stands for Arithmetic Mean - Geometric Mean, and it basically says that for non-negative numbers, the arithmetic mean is always greater than or equal to the geometric mean. Let's put that in math terms:

For non-negative numbers a and b:

(a + b) / 2 ≥ √(ab)

This innocent-looking inequality is surprisingly powerful, and it's going to be our key to unlocking this problem. So, how do we apply it here? Well, let's focus on those terms with the square roots: 2x√(y-1), 2y√(z-1), and 2z√(x-1). We can rewrite these to make the AM-GM inequality more apparent. This step requires mathematical insight and strategic manipulation of expressions.

Applying AM-GM: A Strategic Move

Let's take the first term, 2x√(y-1). We can rewrite this as 2 * √[x²(y-1)]. Now, we can apply the AM-GM inequality to the numbers x² and (y-1). According to AM-GM:

(x² + (y - 1)) / 2 ≥ √[x²(y - 1)]

Multiplying both sides by 2, we get:

x² + (y - 1) ≥ 2√[x²(y - 1)] = 2x√(y - 1)

See where we're going with this? We've managed to create an inequality that looks very similar to the terms in our original problem! We can do the exact same thing for the other two terms. For 2y√(z-1), we get:

y² + (z - 1) ≥ 2y√(z - 1)

And for 2z√(x-1), we get:

z² + (x - 1) ≥ 2z√(x - 1)

Now we have three new inequalities. What's the next step? It's time to add them all up!

Combining the Inequalities: Putting the Pieces Together

If we add the three inequalities we just derived, we get:

[x² + (y - 1)] + [y² + (z - 1)] + [z² + (x - 1)] ≥ 2x√(y - 1) + 2y√(z - 1) + 2z√(x - 1)

Let's simplify this a bit by rearranging the terms:

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

Now, remember our original inequality? It was:

2x√(y - 1) + 2y√(z - 1) + 2z√(x - 1) ≥ xy + yz + zx

We've got a lower bound for the left-hand side of the original inequality. Can we somehow connect this to the right-hand side? This is where another clever trick comes in. We need to relate the sum of squares (x² + y² + z²) to the sum of products (xy + yz + zx). This involves algebraic manipulation and insightful comparison of expressions.

The Final Showdown: Connecting the Dots

Here's another important inequality that might be familiar: For any real numbers x, y, and z:

x² + y² + z² ≥ xy + yz + zx

This is a fundamental inequality that can be proven in various ways (one common way is to consider the sum of squares (x-y)² + (y-z)² + (z-x)² which is always non-negative). Now we're getting somewhere! Let's combine this with our previous inequality:

x² + y² + z² + x + y + z - 3 ≥ 2x√(y - 1) + 2y√(z - 1) + 2z√(x - 1) ≥ xy + yz + zx

So, we have:

x² + y² + z² + x + y + z - 3 ≥ xy + yz + zx

But we also know that x² + y² + z² ≥ xy + yz + zx. This means:

xy + yz + zx + x + y + z - 3 ≥ x² + y² + z² + x + y + z - 3 ≥ xy + yz + zx

The only way this can be true is if all the inequalities become equalities. This is a crucial point in the solution. It significantly narrows down the possibilities and leads us to the final answer. For equality to hold in AM-GM, the terms must be equal. And for equality to hold in x² + y² + z² ≥ xy + yz + zx, we must have x = y = z. Let's unravel what these equalities imply for the original problem.

The Solution Unveiled: Equality Conditions

Let's break down what the equality conditions mean.

Equality in AM-GM: This means that in each application of AM-GM, the terms must be equal. So, we must have x² = y - 1, y² = z - 1, and z² = x - 1.
Equality in x² + y² + z² ≥ xy + yz + zx: This means x = y = z.

Combining these two conditions is the final key. If x = y = z, then our equations x² = y - 1, y² = z - 1, and z² = x - 1 become:

x² = x - 1

This is a simple quadratic equation! Let's solve it:

x² - x + 1 = 0

Using the quadratic formula, we find the discriminant (b² - 4ac) is (-1)² - 4(1)(1) = -3. Since the discriminant is negative, there are no real solutions for x. This might seem like a dead end, but it's actually a crucial piece of the puzzle.

The Twist: A Careful Reassessment

Wait a minute! We made an assumption earlier that needs closer examination. We jumped to the conclusion that equality in the original inequality requires equality in all the AM-GM inequalities and the inequality x² + y² + z² ≥ xy + yz + zx. While this is sufficient for equality in the original inequality, is it necessary? In other words, could there be other solutions where the original inequality holds as an equality, but not all the intermediate inequalities are equalities?

This is a critical point in problem-solving: always re-examine your assumptions! Sometimes, the most elegant solution comes from questioning what seems obvious. Let's go back to the original inequality and think about what it really means for the equality to hold.

The Aha! Moment: A Single Solution

Let's revisit the original inequality:

2x√(y-1) + 2y√(z-1) + 2z√(x-1) ≥ xy + yz + zx

For equality to hold, we need:

2x√(y-1) + 2y√(z-1) + 2z√(x-1) = xy + yz + zx

And we derived that:

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

xy + yz + zx + x + y + z - 3 ≥ x² + y² + z² + x + y + z - 3 ≥ xy + yz + zx

Now, let's think about the specific case where x = y = z = 2. Let's plug these values into the original inequality:

2(2)√(2-1) + 2(2)√(2-1) + 2(2)√(2-1) ≥ (2)(2) + (2)(2) + (2)(2)

4 + 4 + 4 ≥ 4 + 4 + 4

12 = 12

It works! So, (2, 2, 2) is a solution. But is it the only solution? We need to rigorously prove this.

Proving Uniqueness: The Final Step

To prove that (2, 2, 2) is the only solution, let's return to our equality conditions. We know that for equality to hold in the AM-GM inequality, the terms must be equal. So, we need:

x² = y - 1
y² = z - 1
z² = x - 1

If x = y = z, then we already showed that there are no real solutions. So, let's assume that x, y, and z are not all equal. Without loss of generality, let's assume x is the largest of the three. Then:

x ≥ y
x ≥ z

Since x² = y - 1, y² = z - 1, and z² = x - 1, we have:

x² ≥ y² ≥ z²

Taking the square root of each part (remembering that x, y, and z are all at least 1), we get:

x ≥ y ≥ z

But if x ≥ y ≥ z and x² = y - 1, y² = z - 1, and z² = x - 1, then the only way for this to be consistent is if x = y = z. This is a subtle but crucial point. If the variables are not all equal, the inequalities lead to a contradiction. The heart of the argument lies in proving that any deviation from x=y=z=2 will violate the original inequality.

Since we've already shown that x = y = z leads to no real solutions for the quadratic equation x² - x + 1 = 0, the only possibility left is the case where x = y = z = 2. This elegantly demonstrates the uniqueness of the solution.

Conclusion: The Answer and the Journey

So, after all that mathematical detective work, what's the answer? There is only one triplet (x, y, z) of real numbers that satisfies the given inequality: (2, 2, 2).

This problem was a fantastic journey through the world of inequalities. We used the AM-GM inequality, a bit of algebraic manipulation, and some careful reasoning about equality conditions. But perhaps the most important lesson is the power of questioning assumptions and re-examining our approach. Sometimes, the path to the solution isn't a straight line, and it's the twists and turns that make the journey worthwhile. Keep practicing, keep questioning, and keep exploring the beautiful world of mathematics!