So far I have that if $-3 \leq x \leq 4$ then we have that:
$-6 \leq x -3 \leq 1$ and $-1 \leq x+2 \leq 6$
So $|x-3| \leq 6$ and $|x+2| \leq 6$ but I'm not sure how to continue the proof.
2026-04-03 01:05:47.1775178347
How to prove: if $x\in [-3,4]$ then $5\leq |x-3|+|x+2| \leq 7$
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For the first inequality note that $|x+y|\leq |x|+|y|$ so $5=|3-x+x+2|\leq |x-3|+|x+2|$. If $x\geq 3$ you have $|x-3|+|x+2|=2x-1\leq 7$ because $x\leq 4$. If $-2\leq x\leq 3$ you have $|x-3|+|x+2|=-x+3+x+2=5\leq 7$. If $x\leq -2$ you have $|x-3|+|x+2|=-x+3-x-2=-2x+1\leq 7$ because $x\geq -3$.