Absolute value:$ |x-a| \leq \frac{|a|}{2} \Rightarrow |x| \geq \frac{|a|}{2}$ 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!
 A: 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).
A: $i.a\geqslant 0$
$$|x-a|\leqslant\frac a2\Longleftrightarrow\frac a2\leqslant x\leqslant \frac{3a}{2}\Longrightarrow|x|\geqslant\frac{|a|}{2}$$
$ii.a< 0$
$$|x-a|\leqslant-\frac a2\Longleftrightarrow\frac {3a}{2}\leqslant x\leqslant \frac{a}{2}\Longrightarrow|x|\geqslant\frac{|a|}{2}$$
Furthermore, easy to get $$|x|\leqslant{\frac{3|a|}{2}}$$ 
According to No.$i$ and No.$ii$,we can prove it easily. 
A: 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| $.
