Absolute value:$ |x-a| \leq \frac{|a|}{2} \Rightarrow |x| \geq \frac{|a|}{2}$

Question

Prove that

$$|x-a| \leq \frac{|a|}{2} \Rightarrow |x| \geq \frac{|a|}{2}$$

I've tried by direct proof and contradicion but nothing worked. I would like a hint or a tip of what should I do.

Thanks in advance!

Answer 1

Solution:

Let $y = \frac{a}{2} $. Then $a = 2y$ and $|y| = \frac{|a|}{2} $. We can show the contrapositive of the statement:

$$ |x| < |y| \implies |x-2y| > |y| $$

To show this, notice

$$ 3|y| = |3y| = |2y +y +x-x | = |(2y-x) + y + x| \leq |2y - x| + |y| + |x| < |2y-x| + 2|y| $$

Substracting $2|y|$ from both sides gives

$$ |2y - x| > |y| $$

But $|2y - x| = |x - 2y| $.

Answer 2

What kind of proof do you want? Doing this visually on a number line is clear and easy. Even if you do not want to present such a proof, doing this first often gives you an idea on how to do the formal proof. That's what I did to find (one of) my formal proofs for your question.

Now that I am back from church, I can add more detail... but others have already written what I was going to write. The simplest formal proof I came up with was explained by @abel in his comment to the main question: "Use $|a+b|≥||a|−|b||$ with $b=x−a$."

If you cannot use that identity, you can go back to basics. The visual number line proof involves viewing the interval around the point $a$ on the number line with radius $\frac{|a|}2$. The drawing would depend on whether $a$ is positive or negative, so you should do your proof based in these two cases.

The first step , assuming the hypothesis, would be

$$-\frac{|a|}2\le x-a \le \frac{|a|}2$$

Then use the sign of $a$ to show that $\frac a2\le x$ for $a\ge 0$ and $x\le\frac a2$ for $a\le 0$. I see that @Lorence has given the details for this approach. In my opinion, that is the best answer here (and you should have accepted that one... no offense intended to Willie Rosario).

Answer 3

$i.a\geqslant 0$

$$|x-a|\leqslant\frac a2\Longleftrightarrow\frac a2\leqslant x\leqslant \frac{3a}{2}\Longrightarrow|x|\geqslant\frac{|a|}{2}$$

$ii.a< 0$

$$|x-a|\leqslant-\frac a2\Longleftrightarrow\frac {3a}{2}\leqslant x\leqslant \frac{a}{2}\Longrightarrow|x|\geqslant\frac{|a|}{2}$$

Furthermore, easy to get $$|x|\leqslant{\frac{3|a|}{2}}$$

According to No.$i$ and No.$ii$,we can prove it easily.