Question

If \( 1 < x < y < z \), which of the following has the greatest value?

• A. z(x + 1)
• B. z(y + 1)
• C. x(y + z)
• D. y(x + z)
• E. z(x + y)

Hint:

For questions involving inequalities and finding the greatest or least value, plugging in simple, valid numbers is often the quickest and most intuitive method. Choose numbers that are easy to calculate with, like small integers, while respecting the given conditions.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The problem asks to compare the values of five different algebraic expressions, given an inequality that defines the relationship between the variables \(x, y,\) and \(z\).
Step 2: Key Formula or Approach:
There are two common methods to solve this: 1. Plugging in numbers: Choose simple values for \(x, y,\) and \(z\) that satisfy the condition \(1<x<y<z\) and calculate each option. 2. Algebraic comparison: Use the given inequalities to compare the expressions systematically.
Step 3: Detailed Explanation:
Method 1: Plugging in numbers
Let's choose simple integer values that fit the inequality \(1<x<y<z\). For example, let \(x=2\), \(y=3\), and \(z=4\).
Now, we evaluate each option:
(A) \(z(x + 1) = 4(2 + 1) = 4 \times 3 = 12\)
(B) \(z(y + 1) = 4(3 + 1) = 4 \times 4 = 16\)
(C) \(x(y + z) = 2(3 + 4) = 2 \times 7 = 14\)
(D) \(y(x + z) = 3(2 + 4) = 3 \times 6 = 18\)
(E) \(z(x + y) = 4(2 + 3) = 4 \times 5 = 20\)
Based on these values, option (E) has the greatest value (20).
Method 2: Algebraic Comparison
Let's compare the expressions. We are given \(1<x<y<z\). All variables are positive.
Compare (D) and (E):
(D) is \(y(x + z) = yx + yz\)
(E) is \(z(x + y) = zx + zy\)
Since \(yz = zy\), we just need to compare \(yx\) and \(zx\). As \(y<z\) and \(x>0\), it follows that \(yx<zx\). Therefore, \(y(x+z)<z(x+y)\), meaning (E)>(D).
Compare (B) and (E):
(B) is \(z(y + 1)\)
(E) is \(z(x + y)\)
Since \(z\) is a positive common factor, we compare \((y + 1)\) with \((x + y)\). Subtracting \(y\) from both, we compare \(1\) with \(x\). The problem states \(1<x\), so \(1<x\). This implies \((y+1)<(y+x)\). Therefore, \(z(y+1)<z(x+y)\), meaning (E)>(B).
Through these comparisons, we can establish that \(z(x+y)\) is the largest expression. It multiplies the largest number (\(z\)) by the sum of the other two numbers (\(x+y\)), which are both larger than 1.
Step 4: Final Answer:
Both methods show that the expression with the greatest value is \(z(x + y)\).