Question

If x, y, z are distinct positive real numbers the \( (x^2(y+z) + y^2(x+z) + z^2(x+y))/xyz \) would always be

• A. Less than 6
• B. greater than 8
• C. greater than 6
• D. Less than 8

Hint:

Divide every term of the numerator by xyz and group the six ratios in pairs like x/y and y/x. What is the smallest value a number plus its reciprocal can take?

Answer 1

The Correct Option is C

Step 1: Understand the question:
We have three distinct positive numbers x, y, z. We must say what is always true about \(\frac{x^2(y+z) + y^2(x+z) + z^2(x+y)}{xyz}\).

Step 2: Split the fraction into six simple ratios:
Divide each term of the numerator by xyz.
\[ \frac{x^2(y+z)}{xyz} = \frac{x(y+z)}{yz} = \frac{x}{z} + \frac{x}{y} \]
In the same way \(\frac{y^2(x+z)}{xyz} = \frac{y}{z} + \frac{y}{x}\) and \(\frac{z^2(x+y)}{xyz} = \frac{z}{y} + \frac{z}{x}\).
So the expression equals
\[ \left(\frac{x}{y} + \frac{y}{x}\right) + \left(\frac{y}{z} + \frac{z}{y}\right) + \left(\frac{z}{x} + \frac{x}{z}\right) \]

Step 3: Use the fact that a number plus its reciprocal is at least 2:
For any positive a, \(a + \frac{1}{a} - 2 = \frac{(a-1)^2}{a} \ge 0\). So \(a + \frac{1}{a} \ge 2\), and equality holds only when a = 1.
Put \(a = \frac{x}{y}\). Then \(\frac{x}{y} + \frac{y}{x} \ge 2\), with equality only when x = y.
Each of the three brackets is therefore at least 2, so the whole expression is at least 6.

Step 4: Use that the numbers are distinct:
Equality needs x = y, y = z and z = x. The question says x, y, z are distinct, so none of the brackets can equal 2.
Each bracket is strictly greater than 2, so the expression is strictly greater than 6. So it is always greater than 6.

Step 5: Why the other options are wrong:
Option 1 (less than 6) is impossible, because the value is above 6.
Option 2 (greater than 8) is not always true. Take x = 1, y = 1.5, z = 2. The numerator is \(1 \cdot 3.5 + 2.25 \cdot 3 + 4 \cdot 2.5 = 3.5 + 6.75 + 10 = 20.25\) and xyz = 3, so the value is 6.75, which is below 8.
Option 4 (less than 8) is not always true. Take x = 1, y = 2, z = 4. The numerator is \(1 \cdot 6 + 4 \cdot 5 + 16 \cdot 3 = 74\) and xyz = 8, so the value is 9.25, which is above 8.
Only "greater than 6" holds for every choice, which is option 3.

Final Answer:
The expression is a sum of three pairs of the form a + 1/a, so it is at least 6, and strictly more than 6 for distinct numbers. This is option 3. \[ \boxed{\text{greater than } 6 \; \text{(option 3)}} \]