For real numbers $z$ and $w$, $|(1+z)(1+w)-1| \leq (1+|z|)(1+|w|)-1$.

Question

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).

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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 $|(1+z)(1+w) -1| \leq (1 + |z|)(1 + |w|) -1$.

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Answer 1

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

Answer 2

To answer the solution-verification part of the question, this step is wrong.

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}

The above is of the form $\, a \cdot |b| \le c \implies a \cdot b \le c\,$, which does not hold true in general. For example, if $\,a=b=-2, \,c=1\,$ then $\,a \cdot |b| = -4 \le 1 = c\,$, but $\,a \cdot b = 4 \gt 1 = c\,$. The implication does hold true for $\,a \ge 0\,$, but here $\,a = 1 + z\,$ which is not necessarily positive.

The correct way to derive the inequality is to use that $\,|a \cdot b| = |a| \cdot |b|\,$, then:

$$ (1+z)\cdot (1+w) \;\leq\; |(1+z)\cdot(1+w)| \;=\; |1+z|\cdot|1+w| $$