Induction Proof & Inequality: Math Problem Solution

Hey guys! Let's tackle a cool math problem involving induction and inequalities. We've got two parts to this challenge: first, we'll prove a summation formula using mathematical induction, and then we'll dive into proving an inequality in the realm of real numbers. So, buckle up, grab your thinking caps, and let’s get started!

1. Induction Proof: Summation Formula

The first part asks us to demonstrate, using the principle of mathematical induction, that for every positive integer n (denoted as n ∈ N*), the following summation formula holds true:

∑_{k=1}^{n} 1/(k(k+1)(k+2)) = n(n+3) / (4(n+1)(n+2))

This might look a bit intimidating at first, but don't worry! We'll break it down step-by-step. The core idea behind mathematical induction is to prove a statement for all natural numbers by showing it's true for a base case (usually n=1) and then proving that if it's true for some arbitrary number n, it must also be true for n+1. This "domino effect" ensures the statement holds for all n.

1.1. Base Case (n = 1)

Let's start with the base case, where n = 1. We need to show that the formula holds true when we plug in 1 for n. So, the left-hand side (LHS) of the equation becomes:

∑_{k=1}^{1} 1/(k(k+1)(k+2)) = 1/(1(1+1)(1+2)) = 1/(1 * 2 * 3) = 1/6

Now, let's calculate the right-hand side (RHS) of the equation with n = 1:

n(n+3) / (4(n+1)(n+2)) = 1(1+3) / (4(1+1)(1+2)) = 4 / (4 * 2 * 3) = 4/24 = 1/6

As you can see, the LHS (1/6) equals the RHS (1/6), so the formula holds true for the base case n = 1. This is our starting point – the first domino has fallen!

1.2. Inductive Hypothesis

Next, we make the inductive hypothesis. This is a crucial step where we assume that the formula holds true for some arbitrary positive integer n = m. In other words, we assume that:

∑_{k=1}^{m} 1/(k(k+1)(k+2)) = m(m+3) / (4(m+1)(m+2))

This assumption is the cornerstone of our proof. We're saying, "Okay, let's imagine this formula is true for some number m. Now, can we prove it's also true for the next number, m+1?"

1.3. Inductive Step

Now comes the heart of the induction proof: the inductive step. Here, we need to show that if the formula holds true for n = m (our inductive hypothesis), then it must also hold true for n = m+1. In other words, we need to prove that:

∑_{k=1}^{m+1} 1/(k(k+1)(k+2)) = (m+1)(m+4) / (4(m+2)(m+3))

To do this, we'll start with the left-hand side of the equation for n = m+1 and try to manipulate it using our inductive hypothesis to arrive at the right-hand side.

∑_{k=1}^{m+1} 1/(k(k+1)(k+2)) can be rewritten as the sum up to m plus the additional term when k = m+1:

∑{k=1}^{m+1} 1/(k(k+1)(k+2)) = ∑{k=1}^{m} 1/(k(k+1)(k+2)) + 1/((m+1)(m+2)(m+3))

Now, here's where the inductive hypothesis comes into play! We assumed that the sum up to m is equal to m(m+3) / (4(m+1)(m+2)). So, we can substitute that into the equation:

= m(m+3) / (4(m+1)(m+2)) + 1/((m+1)(m+2)(m+3))

Our goal now is to manipulate this expression algebraically to match the right-hand side of the equation for n = m+1, which is (m+1)(m+4) / (4(m+2)(m+3)). This often involves finding a common denominator and simplifying.

Let's find a common denominator, which is 4(m+1)(m+2)(m+3). We get:

= [m(m+3)(m+3) + 4] / [4(m+1)(m+2)(m+3)]

Now, let's expand the numerator:

= [m(m^2 + 6m + 9) + 4] / [4(m+1)(m+2)(m+3)]

= (m^3 + 6m^2 + 9m + 4) / [4(m+1)(m+2)(m+3)]

Now, we need to factor the cubic polynomial in the numerator. Notice that if we plug in m = -1, the numerator becomes zero. This suggests that (m+1) is a factor. Let's perform polynomial division or use synthetic division to divide the numerator by (m+1). We get:

(m^3 + 6m^2 + 9m + 4) = (m+1)(m^2 + 5m + 4)

Now, we can factor the quadratic term:

(m^2 + 5m + 4) = (m+1)(m+4)

So, the numerator becomes:

(m+1)(m+1)(m+4)

Now, let's plug this back into our expression:

= [(m+1)(m+1)(m+4)] / [4(m+1)(m+2)(m+3)]

We can cancel out one (m+1) term from the numerator and denominator:

= [(m+1)(m+4)] / [4(m+2)(m+3)]

And there you have it! We've arrived at the right-hand side of the equation for n = m+1:

(m+1)(m+4) / (4(m+2)(m+3))

This completes the inductive step. We've shown that if the formula holds true for n = m, it also holds true for n = m+1. This is the crucial link in the domino chain!

1.4. Conclusion

By the principle of mathematical induction, we've proven that the formula:

∑_{k=1}^{n} 1/(k(k+1)(k+2)) = n(n+3) / (4(n+1)(n+2))

holds true for all positive integers n (n ∈ N*). We started with the base case, made the inductive hypothesis, and then successfully completed the inductive step. That's one math problem down!

2. Inequality Proof in R^2

Now, let's move on to the second part of our challenge, which involves proving an inequality in the realm of real numbers (R^2). We need to show that for all pairs of real numbers (x, y) ∈ R^2, the following inequality holds true:

|x - y| ≤ 2√(x^2 + y^2 + xy)

This looks like a different beast altogether, but we'll tackle it with a similar methodical approach. The key here is to manipulate the inequality using valid algebraic operations until we arrive at a statement that we know is always true.

2.1. Squaring Both Sides

Since both sides of the inequality are non-negative (the absolute value is always non-negative, and the square root is also non-negative), we can square both sides without changing the direction of the inequality. This often simplifies things by getting rid of the square root and absolute value.

Squaring both sides, we get:

(x - y)^2 ≤ [2√(x^2 + y^2 + xy)]^2

Expanding the left-hand side and simplifying the right-hand side, we have:

x^2 - 2xy + y^2 ≤ 4(x^2 + y^2 + xy)

2.2. Expanding and Rearranging

Now, let's expand the right-hand side and rearrange the inequality to bring all terms to one side:

x^2 - 2xy + y^2 ≤ 4x^2 + 4y^2 + 4xy

0 ≤ 3x^2 + 3y^2 + 6xy

2.3. Dividing by 3

We can divide both sides of the inequality by 3 (since 3 is positive, it doesn't change the direction of the inequality):

0 ≤ x^2 + y^2 + 2xy

2.4. Recognizing a Perfect Square

Now, look closely at the right-hand side. It's a perfect square trinomial! We can rewrite it as:

0 ≤ (x + y)^2

2.5. The Final Step: A True Statement

And there it is! We've arrived at a statement that is always true: the square of any real number (x + y) is always greater than or equal to zero. This is a fundamental property of real numbers.

Since we arrived at a true statement by performing valid algebraic operations on the original inequality, we've proven that the original inequality must also be true.

2.6. Conclusion

We've successfully shown that for all (x, y) ∈ R^2, the inequality:

|x - y| ≤ 2√(x^2 + y^2 + xy)

holds true. We did this by squaring both sides, rearranging terms, recognizing a perfect square, and arriving at a statement that we know is always true.

Conclusion: Math Problem Solved!

Alright, guys! We've conquered both parts of this math challenge. We used mathematical induction to prove a summation formula and algebraic manipulation to prove an inequality in R^2. These are fundamental techniques in mathematics, and mastering them will open doors to tackling more complex problems. Keep practicing, keep exploring, and keep those math muscles strong!