Prove that $\frac1{x}-x$ is multiplicatively subadditive in $(0, 1)$ If
$f(x) = \dfrac1{x}-x$,
prove that
$f(x)$ is multiplicatively subadditive in
$(0, 1)$,
that is,

if $a, b \in (0, 1)$,
then
$f(a)+f(b) &lt; f(ab)$.
To show one expression is larger than another, it is often helpful to consider the difference:

$$\begin{align}
f(ab) - f(a) - f(b) &amp;= \frac{1}{ab} - \frac{1}{a} - \frac{1}{b} + a + b - ab\\
&amp;= \left(\frac{1}{ab} - \frac{1}{a} - \frac{1}{b} + 1\right) - \left(1 - a - b + ab\right)\\
&amp;= \left(\frac{1}{a} - 1\right)\left(\frac{1}{b} - 1\right) - (1-a)(1-b)\\
&amp;= \frac{(1-a)(1-b)}{ab} - (1-a)(1-b)\\
&amp;= (1-a)(1-b)\left(\frac{1}{ab} - 1\right)\\
&amp;&gt; 0.
\end{align}$$
Suppose we take $f(a)$ as $a$ approaches $0$. Then $f(a)$ approaches infinity. If we take $f(b)$ approaching $1$, then $f(b)$ approaches $0$. If we take $f(ab)$ with the same conditions, then ab should approach $0$ - but it is less than $f(a)$, which is important to note, because that means $f(ab)$ approaches infinity faster than $f(a)$ does.