Show $(y/2)/\sqrt{1-x^y}$ is bounded by one for $x\in(0,3/4)$ and $y\in(0,1)$ I would like to show that for all $x\in(0,3/4)$ and $y\in(0,1)$ that $f(x,y) = \frac{y}{2\sqrt{1-x^y}}$ is bounded above by one.
My thought is that for any $x$ I can try to show it an increasing function of $y$ on $(0,1)$. Then it would be bounded by $f(x,1) \leq 1$. To do this I computed the derivative wrt $y$ $\frac{df}{dy} (x,y) = \frac{2(1-x^y)+x^y\log(x^y)}{4 (1-x^y)^{3/2}}$, but I'm not sure how to proceed. I feel like there must be some bound for logs which I can apply here to show the derivative is non-negative.
 A: I think I might have figured it out:
Consider the function
\begin{align*}
    g(x,y) = \frac{y}{2\sqrt{1-x^y}}.
\end{align*}
This function is clearly increasing (thanks @Eric) in $x$ for all $y\in(0,1)$, so it suffices to set $x=3/4$.
Thus, define
\begin{align*}
    f(y) = \log \left( \frac{y}{2\sqrt{1-(3/4)^y}} \right)
\end{align*}
which has derivative
\begin{align*}
    f'(y) = \frac{1}{y} - \frac{\log(4/3)}{2((4/3)^y-1)}.
\end{align*}
We aim to show $f'(y)$ is non-negative.
Note that $(4/3)^y-1 > \log(4/3)y$ for all $y$ so
\begin{align*}
    \frac{\log(4/3)}{2((4/3)^y-1)}
    \leq \frac{\log(4/3)}{2 \log(4/3)y}
    = \frac{1}{2y}.
\end{align*}
Therefore $f'(y) \geq 1/(2y) \geq 0$ so $f(y)$ is monotonic. This implies that $g(3/4,y)$ is also monotonic, so it is bounded by $g(3/4,1) = 1$
A: The function is monotonic increase in both $x$ and $y$ so you need test the corner case $(3/4,1)$, where the value is $1$.

We know the function is monotonic in each variable:

*

*Hold $y$ constant... as $x$ gets bigger, the denominator gets smaller, so the function gets bigger.

*Hold $x$ constant... as $y$ gets bigger, the numerator gets bigger linearly, and the denominator gets larger but less than linearly, and hence the overall function gets bigger.

