Is there a proof for triangle inequality in $\mathbb{R}$ by contradiction/absurd?

Question

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!

Answer 1

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 $|a| + |b| < |a + b|$.

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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, $|a + b| \geq |a + b|$.

Answer 2

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.