For some positive and distinct real numbers $x, y$ and $z$, if $\frac{1}{\sqrt{y}+\sqrt{z}}$ is the arithmetic mean of $\frac{1}{\sqrt{x}+\sqrt{z}}$ and $\frac{1}{\sqrt{x}+\sqrt{y}}$, then the relationship which will always hold true, is

We're told that $\frac{1}{\sqrt{y}+\sqrt{z}}$ is the arithmetic mean of $\frac{1}{\sqrt{x}+\sqrt{z}}$ and $\frac{1}{\sqrt{x}+\sqrt{y}}$.

If $A$ is the arithmetic mean of $B$ and $C$, then $A = \frac{B + C}{2}$

So our condition is:

$\frac{1}{\sqrt{y}+\sqrt{z}} = \frac{1}{2}\left(\frac{1}{\sqrt{x}+\sqrt{z}} + \frac{1}{\sqrt{x}+\sqrt{y}}\right)$

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Multiply both sides by $2$:

$\frac{2}{\sqrt{y}+\sqrt{z}} = \frac{1}{\sqrt{x}+\sqrt{z}} + \frac{1}{\sqrt{x}+\sqrt{y}}$

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To make this easier, let's substitute:

• $a = \sqrt{x}$
• $b = \sqrt{y}$
• $c = \sqrt{z}$

Our equation becomes:

$\frac{2}{b+c} = \frac{1}{a+c} + \frac{1}{a+b}$

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To add the fractions on the right side, find common denominator:

$\frac{2}{b+c} = \frac{(a+b) + (a+c)}{(a+c)(a+b)} = \frac{2a+b+c}{(a+c)(a+b)}$

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Cross multiply:

$2(a+c)(a+b) = (2a+b+c)(b+c)$

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Expand the left side: $2(a+c)(a+b) = 2(a^2 + ab + ac + bc)$

Expand the right side: $(2a+b+c)(b+c) = 2ab + 2ac + b^2 + 2bc + c^2$

Setting them equal:

$2a^2 + 2ab + 2ac + 2bc = 2ab + 2ac + b^2 + 2bc + c^2$

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Cancel $2ab + 2ac + 2bc$ from both sides:

$2a^2 = b^2 + c^2$

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Substitute back $a = \sqrt{x}$, $b = \sqrt{y}$, $c = \sqrt{z}$:

$2(\sqrt{x})^2 = (\sqrt{y})^2 + (\sqrt{z})^2$

$2x = y + z$

Therefore, the relationship that always holds true is $2x = y + z$.