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I want to prove that given $a,b,c\in\mathbb{R}$ we have $|a+b|\leq|a|+|b|$ using an absurd and reaching a contradiction.

So, I state, by absurd, that $|a+b|>|a|+|b|$, but I can't reach the contradiction. It look simple, but I'm afraid that it isn't. In my research I didn't find this proof. Thanks for any contribution!

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    $\begingroup$ Here's a proof by contradiction. Assume $|a+b| > |a|+|b|$. --Insert here a proof of the triangle inequality--, then $|a+b| \le |a|+|b|$, a contradiction. $\endgroup$
    – jjagmath
    Commented Nov 26, 2021 at 16:01
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    $\begingroup$ @jjagmath By proof by contradiction, we generally mean proofs that aren't easily reworked into direct proofs. $\endgroup$
    – Rushabh Mehta
    Commented Nov 26, 2021 at 16:31
  • $\begingroup$ p.s. In a proof by contradiction, the absurdity isn't that <initial negation of the desired result>, but rather, the derived contradiction itself. And this contradiction need not directly involve either the desired result or its negation. $\endgroup$
    – ryang
    Commented Nov 26, 2021 at 16:39
  • $\begingroup$ @DonThousand From the logic point of view, it IS a proof by contradiction. You may argue that it a very artificial one, but so is trying to fit a proof method to a theorem. $\endgroup$
    – jjagmath
    Commented Nov 26, 2021 at 19:43

2 Answers 2

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It depends on what you already know about absolute values and inequalities.

For instance, have you already shown that if $a > b > 0$ and $c > 0$, then $ac > bc$? From that we get $a^2 > ab$ and $ab > b^2$, so $a^2 > b^2$. Also for any $a,b, |a|^2 = a^2$, $|a||b| = |ab|$, and $a \le |a|$.

Thus if $$|a + b| > |a| + |b|\\|a + b|^2 >(|a| + |b|)^2\\(a + b)^2 > (|a| + |b|)^2\\a^2 + 2ab + b^2 > a^2 + 2|ab| + b^2\\ab > |ab|$$ which is absurd.

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Our claim is that $|a| + |b| \geq |a + b|$ for real $a$ and $b$ and we want to prove it by contradiction. Hence, we assume that $|a| + |b| < |a + b|$.

Here is a proof.

Case 1: $0 < a < b$.

Because both $a$ and $b$ are greater than zero, this means that $a + b$ is also greater than zero. Then $|a| = a$, $|b| = b$, and $|a + b| = a + b$. We then have $a + b > a + b$ which is a contradiction.

Case 2: $a < 0 < b$ and $|a| < |b|$.

This means that $|a| + |b| > 0$. Because of the conditions, we have $b > a + b \implies |b| > |a + b|$. Since we have $|a| + |b| > 0$, then $|a| + |b| > |b| > |a + b| \implies |a| + |b| > |a + b|$. Contradiction.

Case 3: $a < 0 < b$ and $|a| > |b|$.
This also means that $|a| + |b| > 0$. Because of the conditions, $a < a + b \implies |a| > |a + b|$. Since we have $|a| + |b| > 0$, then $|a| + |b| > |a| > |a + b| \implies |a| + |b| > |a + b|$. Contradiction.

Case 4: $a < b < 0$.

This case can be treated in a similar manner as the first case. Since both $a$ and $b$ are less than zero, this means that $a + b$ is also less than zero. Then $|a| = -a$, $|b| = -b$, and $|a + b| = -a - b$ and we have $-a - b > -a - b$. Contradiction.

Case 5: $a = b$. This means that $|2a| < |2a|$ which is a contradiction.

Case 6: $a = 0$. This means that $|b| < |b|$ which is a contradiction.

Because all cases of $a$ and $b$ lead to a contradiction, our assumption must not be true. Therefore, $|a + b| \geq |a + b|$.

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  • $\begingroup$ There's also no reason an inequality (involving negatives and positives!) without absolute value signs remains true with absolute value signs, so be careful about this, too. Some cases are required. $\endgroup$
    – Jakob Streipel
    Commented Nov 26, 2021 at 16:16
  • $\begingroup$ @prets Isn't it that one case is enough to prove the claim by contradiction? $\endgroup$
    – soupless
    Commented Nov 26, 2021 at 16:26
  • $\begingroup$ Sure, but all you're assuming in your original assumption is that for some $a$ and $b$, we have $|a| + |b| < |a + b|$. You've no idea which, and how their sizes compare, etc. $\endgroup$
    – Jakob Streipel
    Commented Nov 26, 2021 at 16:27
  • $\begingroup$ @prets This means that I should: 1) state which values of $a$ and $b$ make the equation true, and/or; 2) state all the possible cases and prove them one by one, right? $\endgroup$
    – soupless
    Commented Nov 26, 2021 at 16:32
  • $\begingroup$ Yes. Since you know nothing about the assumed $a$ and $b$, you would need to show that whatever they are, they lead to a contradiction. $\endgroup$
    – Jakob Streipel
    Commented Nov 26, 2021 at 16:33

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