Let $f: \mathbb R\longrightarrow\mathbb R^{+}$ be a non-decreasing function. Show that there is a point $a\in\mathbb R$ that
$$f\left(a+\frac1{f(a)}\right) \lt 2f(a).$$

Suppose to the contrary that $f(a+\frac1{f(a)})\ge 2f(a)$ for all $a$. Define a sequence of real numbers $x_n$ by $x_0=0$ and $x_n = x_{n-1}+\frac{1}{f(x_{n-1})}$. Our assumption implies $f(x_n)\ge 2f(x_{n-1})$ for all $n$. Iterating this inequality $n$ times yields $f(x_n)\ge 2^nf(x_0)$, so $f(x_{n})$ grows without bound as $n\to\infty$.

On the other hand, since $f(x_{n-1})\ge 2^{n-1}f(x_0)$, we have $x_n= x_{n-1}+\frac1{f(x_{n-1})}\le x_{n-1}+\frac1{f(x_0)2^{n-1}}$ for all $n$. Iterating this inequality $n$ times yields
$$
x_n \le x_0 + \frac1{f(x_0)2^{0}}+ \frac1{f(x_0)2^{1}}+\dots+ \frac1{f(x_0)2^{n-1}}=0+\frac1{f(x_0)}(2^{-0}+2^{-1}+\dots 2^{-(n-1)})\le \frac2{f(x_0)}
$$
Therefore, since $f$ is non-decreasing, we have $f(x_n)\le f(\frac2{f(x_0)})$, contradicting the unboundedness of $f(x_{n})$.

Suppose to the contrary that $f(a+\frac1{f(a)})\ge 2f(a)$ for all $a$. Define a sequence of real numbers $x_n$ by $x_0=0$ and $x_n = x_{n-1}+\frac{1}{f(x_{n-1})}$. Our assumption implies $f(x_n)\ge 2f(x_{n-1})$ for all $n$. Iterating this inequality $n$ times yields $f(x_n)\ge 2^nf(x_0)$, so $f(x_{n})$ grows without bound as $n\to\infty$.

On the other hand, since $f(x_{n-1})\ge 2^{n-1}f(x_0)$, we have $x_n= x_{n-1}+\frac1{f(x_{n-1})}\le x_{n-1}+\frac1{f(x_0)2^{n-1}}$ for all $n$. Iterating this inequality $n$ times yields $$ x_n \le x_0 + \frac1{f(x_0)2^{0}}+ \frac1{f(x_0)2^{1}}+\dots+ \frac1{f(x_0)2^{n-1}}=0+\frac1{f(x_0)}(2^{-0}+2^{-1}+\dots 2^{-(n-1)})\le \frac2{f(x_0)} $$ Therefore, since $f$ is non-decreasing, we have $f(x_n)\le f(\frac2{f(x_0)})$, contradicting the unboundedness of $f(x_{n})$.