<s>[INST] For real numbers $z$ and $w$, $|(1+z)(1+w)-1| \leq (1+|z|)(1+|w|)-1$. I have written an attempted proof of the theorem on the title, and I need help verifying it.
I have used the following theorems to proof the theorem on the title.
Theorem 5.14)a) Let $x$ be a real number. $-|x| \leq x \leq |x|$.
Theorem 5.14)b) Let $a \geq 0$. $|x| \leq a$ if and only if $-a \leq x \leq a$.
Theorem 5.14)c) Let $x$ and $y$ be real numbers. $|x+y| \leq |x| + |y|$ (The Triangle Inequality).

For real numbers $z$ and $w$, $|(1+z)(1+w)-1| \leq (1+|z|)(1+|w|)-1$.
Proof. From Theorem 5.14)a), $(1+z) \leq |(1+z)|$ and $(1+w) \leq |(1+w)|$. Multiplying the $(1+z) \leq |(1+z)|$ by $|(1+w)|$, one obtains
\begin{align}
(1+z)|(1+w)| \leq |(1+z)||(1+w)|
\end{align}
Since $(1+w) \leq |(1+w)|$, it follows that
\begin{align}
(1+z)(1+w) \leq |(1+z)||(1+w)|
\end{align}
Observe that from The Triangle Inequality,
\begin{align}
|(1+z)| \leq 1 + |z|\\
|(1+w)| \leq 1 + |w|
\end{align}
Since $|(1+z)|(1 + |w|) \leq (1 + |z|)(1 + |w|)$ and $|(1+w)| \leq (1 + |w|)$, it follows that
\begin{align}
(1+z)(1+w) -1 \leq |(1+z)||(1+w)| -1 \leq (1 + |z|)(1 + |w|) -1
\end{align}
It is clear that $(1 + |z|)(1 + |w|) -1 \geq 0$.
In case that $(1+z)(1+w) -1 \geq 0$, $|(1+z)(1+w) -1| = (1+z)(1+w) -1$.
Hence,
\begin{align}
-[(1 + |z|)(1 + |w|) -1] \leq 0 \leq (1+z)(1+w) -1 \leq (1 + |z|)(1 + |w|) -1
\end{align}
From Theorem 5.14)b),
\begin{align}
|(1+z)(1+w) -1| \leq (1 + |z|)(1 + |w|) -1
\end{align}
establishing the result for this case.
On the other hand, in case that $(1+z)(1+w) -1 < 0$, $|(1+z)(1+w) -1| = -[(1+z)(1+w) -1]$.
Hence,
\begin{align}
-[(1 + |z|)(1 + |w|) -1] \leq (1+z)(1+w) -1 < 0
\end{align}
Since $(1 + |z|)(1 + |w|) -1 \geq 0$,
\begin{align}
-[(1 + |z|)(1 + |w|) -1] \leq (1+z)(1+w) -1 \leq (1 + |z|)(1 + |w|) -1
\end{align}
Multiplying the inequalities by $-1$, one obtains
\begin{align}
(1 + |z|)(1 + |w|) -1 \geq -[(1+z)(1+w) -1] \geq -[(1 + |z|)(1 + |w|) -1]
\end{align}
From Theorem 5.14)b),
\begin{align}
|(1+z)(1+w) -1| \leq (1 + |z|)(1 + |w|) -1
\end{align}
establishing the result for this case.
Because the result for all cases of $(1+z)(1+w) -1$ have been established, it is the case that $|(1+z)(1+w) -1| \leq (1 + |z|)(1 + |w|) -1$.

 [/INST] We have :
$(1+z)(1+w)-1 = z+w+zw$
Thus :
$|(1+z)(1+w)-1| = |z+w+zw| \le |z| + |w| + |zw| = (1+|z|)(1+|w|) -1$
QED
</s>