For some positive and distinct real numbers x,y and z,ify+z1 is the arithmetic mean of \frac{1}{\x + \z}and \frac{1}{\\
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Official Correct Answer: A. The question says: For some positive and distinct real numbers x, y, and z, if 1 over (sqrt(y) + sqrt(z)) is the arithmetic mean of 1 over (sqrt(x) + sqrt(z)) and 1 over (sqrt(x) + sqrt(y)), then which relationship always holds true? The options are about arithmetic or geometric progressions or something related to the square roots being in progression.
Okay, so first, let's parse the given condition. The arithmetic mean (AM) of two numbers is just the average of those two numbers. So, if I have two numbers, say a and b, their AM is (a + b)/2. In this case, the AM is given as 1/(sqrt(y) + sqrt(z)), and it's the average of 1/(sqrt(x) + sqrt(z)) and 1/(sqrt(x) + sqrt(y)).
So, mathematically, this translates to:
1/(sqrt(y) + sqrt(z)) = [1/(sqrt(x) + sqrt(z)) + 1/(sqrt(x) + sqrt(y))]/2
Hmm, that's a bit complex, but I can work with it. Let me write that equation down:
$$
\frac{1}{\sqrt{y} + \sqrt{z}} = \frac{1}{2} \left( \frac{1}{\sqrt{x} + \sqrt{z}} + \frac{1}{\sqrt{x} + \sqrt{y}} \right)
$$
Alright, so my goal is to manipulate this equation to find a relationship between x, y, and z. Simplifying this step by step.
First, multiply both sides by 2 to eliminate the denominator on the right:
$$
\frac{2}{\sqrt{y} + \sqrt{z}} = \frac{1}{\sqrt{x} + \sqrt{z}} + \frac{1}{\sqrt{x} + \sqrt{y}}
$$
Now, let's denote sqrt(x) as a, sqrt(y) as b, and sqrt(z) as c for simplicity. So, a = sqrt(x), b = sqrt(y), c = sqrt(z). Then, the equation becomes:
$$
\frac{2}{b + c} = \frac{1}{a + c} + \frac{1}{a + b}
$$
This substitution might make the algebra a bit cleaner. So, now we have:
$$
\frac{2}{b + c} = \frac{1}{a + c} + \frac{1}{a + b}
$$
Compute the right-hand side (RHS):
$$
\frac{1}{a + c} + \frac{1}{a + b} = \frac{(a + b) + (a + c)}{(a + c)(a + b)} = \frac{2a + b + c}{(a + c)(a + b)}
$$
So, substituting back into the equation:
$$
\frac{2}{b + c} = \frac{2a + b + c}{(a + c)(a + b)}
$$
Now, cross-multiplying to eliminate the denominators:
$$
2 \cdot (a + c)(a + b) = (2a + b + c)(b + c)
$$
Expanding both sides.
First, the left-hand side (LHS):
$$
2(a + c)(a + b) = 2[(a)(a) + a b + a c + b c] = 2[a^2 + a b + a c + b c]
$$
Which simplifies to:
$$
2a^2 + 2a b + 2a c + 2b c
$$
Now, the right-hand side (RHS):
$$
(2a + b + c)(b + c) = 2a(b + c) + (b + c)(b + c)
$$
Computing each part:
First term: 2a(b + c) = 2a b + 2a c
Second term: (b + c)^2 = b^2 + 2b c + c^2
So, adding them together:
2a b + 2a c + b^2 + 2b c + c^2
So, the RHS is:
$$
2a b + 2a c + b^2 + 2b c + c^2
$$
Now, let's set LHS equal to RHS:
2a^2 + 2a b + 2a c + 2b c = 2a b + 2a c + b^2 + 2b c + c^2
Hmm, let's subtract the RHS from both sides to bring everything to one side:
2a^2 + 2a b + 2a c + 2b c - (2a b + 2a c + b^2 + 2b c + c^2) = 0
Simplify term by term:
2a^2 remains.
2a b - 2a b = 0
2a c - 2a c = 0
2b c - 2b c = 0
Now, subtract the remaining terms:
- b^2 - c^2 = 0
So, putting it all together:
2a^2 - b^2 - c^2 = 0
Which can be rewritten as:
2a^2 = b^2 + c^2
Hmm, interesting. So, 2a^2 = b^2 + c^2.
But remember, a, b, c are sqrt(x), sqrt(y), sqrt(z). So, let's substitute back:
2(sqrt(x))^2 = (sqrt(y))^2 + (sqrt(z))^2
Simplify sqrt squared:
2x = y + z
So, 2x = y + z.
Wait, that's a key equation. So, 2x = y + z.
Which implies that x is the average of y and z. So, x is the arithmetic mean of y and z.
But looking at the options, Option A: y, x, z are in arithmetic progression.
Option B: x, y, z are in geometric progression.
Option C: sqrt(x), sqrt(z), sqrt(y) are in arithmetic progression.
Option D: sqrt(x), sqrt(z), sqrt(y) are in arithmetic progression. Wait, actually, looking back, the options might have a typo. Let me check.
Wait, the original options are:
A) y, x, and z are in arithmetic progression
B) x, y, and z are in geometric progression
C) sqrt(x), sqrt(z), and sqrt(y) are in arithmetic progression
D) sqrt(x), sqrt(z), and sqrt(z) are in arithmetic progression
Wait, D seems to have a typo because