Let $A$ and $B$ be two fixed positive real numbers. The function $f$ is defined by
$$
f(x, y)=\min \left\{x, \frac{A}{y}, y+\frac{B}{x}\right\},
$$
for all pairs $(x, y)$ of positive real numbers.
Determine the largest possible value of $f(x, y)$.
We show that the largest possible value of $f(x, y)$ is $\sqrt{A+B}$.
Solution 1.
Let us first determine the pair $(x, y)$ for which $x=\frac{A}{y}=y+\frac{B}{x}$. Solving this gives $x=\sqrt{A+B}$ and $y=\frac{A}{\sqrt{A+B}}$. Clearly, in this case, $f(x, y)=\sqrt{A+B}$, and we claim that this is indeed also the maximum value of $f(x, y)$. The first quadrant can be divided into three (non-disjoint) regions:
I. $00$. In this case it is clear that $f(x, y) \leq x \leq \sqrt{A+B}$.
II. $x>0$ and $y \geq \frac{A}{\sqrt{A+B}}$. Here, $f(x, y) \leq \frac{A}{y} \leq \frac{A}{\sqrt{A+B}}=\sqrt{A+B}$.
III. $x \geq \sqrt{A+B}$ and $0<y \leq \frac{A}{\sqrt{A+B}}$. Here we have $f(x, y) \leq y+\frac{B}{x} \leq \frac{A}{\sqrt{A+B}}+\frac{B}{\sqrt{A+B}}=$ $\sqrt{A+B}$.

Our claim is therefore valid.
Solution 2.
Let $g: \mathbb{R}^{+} \times \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$be the function given by letting $y=\frac{A}{z}$, so that
$$
g(x, z)=f\left(x, \frac{A}{z}\right)=\min \left(x, z, \frac{A}{z}+\frac{B}{x}\right) .
$$

Then $g$ has the same range as $f$, because $z \mapsto \frac{A}{z}$ is a bijecton $\mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$. Now observe that if $x \leq z$ then $g(x, z) \leq \min \left(x, \frac{A}{x}+\frac{B}{x}\right)={ }^{z} g(x, x)$, while if $x \geq z$ then $g(x, z) \leq$ $\min \left(z, \frac{A}{z}+\frac{B}{z}\right)=g(z, z)$.
So we wish to find the maximum of the function $h(x)=g(x, x)=\min \left(x, \frac{A}{x}+\frac{B}{x}\right)$. This must occur when $x=\frac{A+B}{x}$, since $x$ is increasing while $\frac{A+B}{x}$ is decreasing. This gives $x=\sqrt{A+B}$, and we conclude that $\max f(x)=\sqrt{A+B}$.

Solution 3.
We use the fact that $\min (u, v) \leq \sqrt{u v}$ for all $u, v \in \mathbb{R}^{+}$to reduce the problem to just two cases. If $x \leq \frac{A}{y}$, then
$$
\begin{aligned}
f(x, y)=\min \left(x, y+\frac{B}{x}\right) & \leq \min \left(x, \frac{A}{x}+\frac{B}{x}\right) \\
& \leq \sqrt{x\left(\frac{A}{x}+\frac{B}{x}\right)}=\sqrt{A+B} ;
\end{aligned}
$$
while if $x \geq \frac{A}{y}$, then
$$
\begin{aligned}
f(x, y)=\min \left(\frac{A}{y}, y+\frac{B}{x}\right) & \leq \min \left(\frac{A}{y}, y+\frac{B y}{A}\right) \\
& \leq \sqrt{\frac{A}{y}\left(y+\frac{B y}{A}\right)}=\sqrt{A+B} .
\end{aligned}
$$

Hence $f(x, y) \leq \sqrt{A+B}$ for all $x, y$. Moreover, as seen above taking $x=\sqrt{A+B}$, $y=\frac{A}{\sqrt{A+B}}$ we obtain $f(x, y)=\sqrt{A+B}$, so this is the largest possible value of $f$.