Distribution of $Y = \frac{1}{X} + X$ As in the title, I am trying to get the distribution of $Y = \frac{1}{X} + X$, where $X$ is uniformly distributed on the interval (0, 1).  I am having trouble seeing how to move beyond what (very little) I have so far:
$$P(Y\leq y) = P(\frac{1}{X} + X \leq y) = P(\{x: \frac{1}{x} + x \leq y\}) \\
= P(\{ 1 + x^2 \leq xy \}) = P(\{ x^2 - xy + 1 = 0\})
$$
 A: The function $x+\frac{1}{x}$ is decreasing on $(0,1)$, so it's enough to find where $x+\frac{1}{x}=y$, since then $P(Y<y)$ is equal to $1$ minus that value.  We can find that value from the quadratic formula
$$
\frac{y-\sqrt{y^2-4}}{2}
$$
So, $P(Y<y)=1-\frac{y-\sqrt{y^2-4}}{2}$ for $y>2$ and $0$ for $y\leq2$
A: \begin{align}
& \qquad x + \frac 1 x = y \\[6pt]
\Longleftrightarrow & \qquad x^2 + 1 = xy \\[6pt]
\Longleftrightarrow & \qquad x^2 - yx + 1 = 0 & \text{(a quadratic equation)} \\[6pt]
\Longleftrightarrow & \qquad x = \frac{y \pm \sqrt{y^2 - 4}} 2
\end{align}
We need $0<x<1$, so we choose $\text{“}{-}\text{''}$ rather than $\text{“}{+}\text{''}$.
No if you can show that $x+ \dfrac 1 x$ decreases as $x$ increases from $0$ to $1$, that deals with the inequality $(1)$ below. And you can do that by showing its derivative is negative when $0<x<1$.  That implies
$$
X + \frac 1 X \le y \text{ if and only if } X \ge \frac{y - \sqrt{y^2-4}} 2. \tag 1
$$
The probability of that is
$$ 1 - \frac{y - \sqrt{y^2-4}} 2.$$
That gives you the c.d.f. and if you need the p.d.f. you can differentiate.
