A relation $$ is defined on the set $\mathbb{R}^{+}$ of positive real numbers by $$ if the arithmetic mean of $x$ and $y$ equals the geometric mean of $$ and $$, that is, if $$\frac{1}{2} \cdot ( + ) = \sqrt{xy}$$
In order to prove $R$ is an equivalence equation I proved that $R$ is reflexive and symmetric but I am having trouble proving it being transitive.
I supposed $x,y,z \in \mathbb{R}^{+}$ such that $xRy$ and $yRz$ and got two equations
$$ \begin{cases} \frac{1}{2}(x+y)=\sqrt{xy} \\ \frac{1}{2}(y+z)=\sqrt{yz} \\ \end{cases} $$
but I am struggling to reach $\frac{1}{2}( + z) = \sqrt{z}$, I tried multiplying $\sqrt{z}$ and $\sqrt{x}$ to both side of each equations and substituting but I can't seem to eliminate $y$.
Think about what the relation actually means. Can you find a nontrivial example of two numbers whose arithmetic and geometric means match?
It turns out that the relation isn't actually all that interesting: it's just regular equality. To see this, choose any $x, y > 0$ and observe that: \begin{align*} \frac{x + y}{2} = \sqrt{xy} &\iff x + y = 2\sqrt{xy} \\ &\iff (x + y)^2 = (2\sqrt{xy})^2 \\ &\iff x^2 + 2xy + y^2 = 4xy \\ &\iff x^2 - 2xy + y^2 = 0 \\ &\iff (x - y)^2 = 0 \\ &\iff x - y = 0 \\ &\iff x = y \end{align*}
So the given relation is transitive because equality is transitive.