<s>[INST] Prove this inequality: If $|x+3|< 0.5$, show that $|4x+13| < 3$ If $|x + 3| < 0.5$, show that $|4x + 13| < 3$
This is what I've got so far:
$|4x + 13| = |(x + 3) + (3x + 10)|$
by the Triangle Inequality: 
$|(x + 3) + (3x + 10)| \le |x + 3| + |3x + 10|$
Now I continue to apply the Triangle Inequality to reach: 
$|(x + 3) + (3x + 10)| \le                                                           |x + 3| + |x + 3| + |x + 3| + |x + 3| + |1|$
So I come to the conclusion: $|4x + 13| \le 4|x + 3| + 1$
Since $|x + 3| < 0.5$, then $4|x + 3| + 1 < 4\cdot 0.5 + 1 = 3$
Then $|4x + 13| < 3$
Please take a look and let me know if there are errors, if so, enlighten me.
Thank you :)
 [/INST] Yes, your solution is correct. A quicker approach could go like this: $|4x+13|=|4(x+3)+1|\leq 4|x+3|+1<4\cdot\frac12+1=3$.
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