How to prove |a-b|≤|a|+|b|?

Question

I am struggling with

$$ |a-b|\leq |a|+|b|$$

Can anyone please help? Many thanks!

Answer 1

I know that the given inequality can be proved simply as the guys commented above, but I liked to share with you a nice way used in proofs, that is "PROOF BY CASES".

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Consider $6$ cases;

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CASE(1): $a \ge 0,b \ge 0,a \ge b$

We have: $|a-b|=a-b$, we have: $|a|=a$, and we have: $|b|=b$

So, the given inequality, $|a-b| \le |a|+|b|$ becomes $a-b \le a+b$ which reduces to $-b \le b$ which is true since $b$ is positive.

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CASE(2): $a \ge 0,b \ge 0,a \le b$

We have: $|a-b|=b-a$, we have: $|a|=a$, and we have: $|b|=b$

So, the given inequality, $|a-b| \le |a|+|b|$ becomes $b-a \le a+b$ which reduces to $-a \le a$ which is true since $a$ is positive.

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CASE(3): $a \le 0, b \ge 0$

We have: $|a-b|=b-a$, we have: $|a|=-a$, and we have: $|b|=b$

So, the given inequality, $|a-b| \le |a|+|b|$ becomes $b-a \le -a+b$ which is clearly true.

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CASE(4): $a \le 0, b \le 0, a \ge b$

We have: $|a-b|=a-b$, we have: $|a|=-a$, and we have: $|b|=-b$

So, the given inequality, $|a-b| \le |a|+|b|$ becomes $a-b \le -a-b$ which reduces to $a \le -a$ which is true since $a$ is negative.

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CASE(5): $a \le 0, b \le 0, a \le b$

We have: $|a-b|=b-a$, we have: $|a|=-a$, and we have: $|b|=-b$

So, the given inequality, $|a-b| \le |a|+|b|$ becomes $b-a \le -a-b$ which reduces to $b \le -b$ which is true since $b$ is negative.

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CASE(6): $a \ge 0, b \le 0$

We have: $|a-b|=a-b$, we have: $|a|=a$, and we have: $|b|=-b$

So, the given inequality, $|a-b| \le |a|+|b|$ becomes $a-b \le a-b$ which is clearly true.

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All cases are considered above, and there is no more cases to be considered. Therefore, the given inequality always holds true.

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Hope this helps.

Answer 2

$|a-b| = |a+(-b)| \leq |a| + |-b| = |a| + |b|$.