If $\vert x - a \vert < \frac{1}{2}\vert a\vert$, then $\frac{1}{2}\vert a\vert < \vert x \vert$?

Question

I'm reading a proof, and I cannot figure out why the following is true:

If $\vert x - a \vert < \frac{1}{2}\vert a\vert$, then $\frac{1}{2}\vert a\vert < \vert x \vert$

(x and a are reals, of course). I've tried using the triangle inequality every which way, but I just can't prove the result. Please prove that the statement is correct.

Edit: Apologies. Changed the first ">" to "<".

Answer 1

$\vert x-a\vert<\frac12\vert a\vert$ is equivalent with $$a-\frac12\vert a\vert < x<a+\frac12\vert a \vert.$$ If $a>=0$ then $\vert a\vert= a$ so $$\frac12\vert a\vert<x<\frac32\vert a\vert.$$ If $a<0$ then $\vert a\vert= -a$ so $$\frac32a<x<\frac12 a.$$ And hence $x<0$, so $\vert x\vert = -x$. Thus $$\frac12\vert a\vert<\vert x\vert<\frac32\vert a\vert.$$

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Hope this helps.

Answer 2

For convenience, suppose that a is positive. Then we can rewrite the first statement as

-a/2 < x - a < a/2

so that

a/2 < x < 3a/2

The left hand side here is what we want.

Answer 3

This will help immensely! I present the reverse triangle inequality: For all $x,y\in\mathbb{R}$, $|x-y| \geq |x|-|y|$. Notice that since $|x-y|=|y-x|$, this is equivalent to saying $|x-y| \geq |y|-|x|$.

Proof: The triangle inequality says $|a+b| \leq |a|+|b|$ for all real $a,b.$ Apply this to $a = y-x$ and $b = x$. Then, subtract $|a|$ from both sides of the inequality. Finally, use that $|a|=|-a|$.

How this helps: we have $|a|/2 > |x-a| \geq |a|-|x|$. Then, add $|x|$ to both sides and subtract $|a|/2.$

Answer 4

If $a\ne 0$ then we have:

Remember that $$a^2\leq b^2\iff |a|\leq |b|$$ so we might square first inequality:

$$(x - a)^2< \frac{1}{4}a^2$$ then $$(2x-3a)(2x-a)<0$$ so, if $a>0$ then $$x\in ({a\over 2},{3a\over 2})\implies |x|>{|a|\over 2}$$

and if $a<0$ then $$x\in ({3a\over 2},{a\over 2})\implies |x|>{|a|\over 2}$$