# How does one prove the inequality $1+|x|\le (1+|y|)(1+|x-y|)$?

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?)

• Show your work please and you may use that $|x-y| \le |x| +|y|$ or $|x-y| \ge ||x|-|y||$ – Fareed AF Jan 30 at 8:13

From the triangle inequality,

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

thus:

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

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 $$1+|x|\leq (1+ |y|)(1+|x-y|).$$

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

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