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I am trying to understand the proof of a proposition regarding Fourier transform in Wolff's Lecture notes on Harmonic Analysis (see Proposition 1.4 in the linked notes):

Suppose that $f$ is $C^N$ and that $D^\alpha f\in L^1$ for all $\alpha$ with $0\le|\alpha|\le N$. Then $$ \widehat{D^\alpha f}(\xi)=(2\pi i \xi)^\alpha\hat{f}(\xi) $$ when $|\alpha|\le N$ and furthermore $$ |\hat{f}(\xi)|\le C(1+|\xi|)^{-N} $$ for a suitable constant $C$.

In the last step of the proof, the following inequality is used without a proof:

$$ 1+|x|\le (1+|y|)(1+|x-y|), \quad x,y\in {\mathbb R}^n. $$

(See the inequality on page 6 on the linked notes.)

Could anyone show why the inequality above is true? (Does it has some geometric explanation?)

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    $\begingroup$ Show your work please and you may use that $|x-y| \le |x| +|y|$ or $|x-y| \ge ||x|-|y||$ $\endgroup$ – Fareed AF Jan 30 at 8:13
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From the triangle inequality,

$\vert x \vert = \vert x - y + y \vert \le \vert x - y \vert + \vert y \vert; \tag 1$

thus:

$1 + \vert x \vert \le 1 + \vert x - y \vert + \vert y \vert \le 1 + \vert x - y \vert + \vert y \vert + \vert y \vert \vert x - y \vert = (1 + \vert y \vert)(1+ \vert x - y \vert). \tag 2$

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Take the origin and $x,y$ as a triangle in $R^n$, then from triangle inequality we know that $$ |x| \leq |y|+|x-y|, $$ then we have $$1+|x|\leq 1+ |y|+|x-y|+|y||x-y|,$$ that is $$1+|x|\leq (1+ |y|)(1+|x-y|).$$

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This is really just a variant on the answer by Robert Lewis, but the wrinkle I think'll be helpful. If you replace $x$ with $x+y$ (which is OK since the point is to prove the inequality for all $x,y\in\mathbb{R}^n$), the inequality to prove becomes symmetric in $x$ and $y$:

$$1+|x+y|\le(1+|y|)(1+|x|)$$

and this follows from the triangle inequality $|x+y|\le|x|+|y|$, so that

$$1+|x+y|\le1+|x|+|y|\le1+|x|+|y|+|x||y|=(1+|y|)(1+|x|)$$

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