How would I go about proving that

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

Using the triangle inequality?

I tried squaring both sides, yielding:

$$|a|^2 +|b|^2 +2|a||b| \leq 2|a|^2+2|b|^2+2||a|^2-|b|^2|$$

Is it correct to move the terms to the other side? I tried $$2|a||b| \leq |a|^2+|b|^2+2||a|^2-|b|^2|$$ Then it is obvious that $$0\leq |a|^2+|b|^2-2|a||b| +2||a|^2-|b|^2|$$ $$0\leq (|a|-|b|)^2 +2||a|^2-|b|^2|$$ $$0\leq ||a|-|b||^2 +2||a|^2-|b|^2|$$

But I have neither used the triangle inequality, nor it looks mathematically rigorous for my real analysis class. Any tips and tricks for solving these problems?


Hint: Note that




$$2|a|+2|b|\leq 2(|a+b|+|a-b|)$$ then $$|a|+|b|\leq |a+b|+|a-b|$$

  • $\begingroup$ This is a great proof, but you're not using the triangle inequality? $\endgroup$ – George Oct 22 '14 at 17:06
  • 1
    $\begingroup$ @George Yes is used in the fact that $|a+b+a-b|\leq|a+b|+|a-b|$ and $|b+a+b-a|\leq|b+a|+|b-a|$ $\endgroup$ – AsdrubalBeltran Oct 22 '14 at 17:13

An inelegant solution uses cases:

If $a \ge 0$ and $b \ge 0$ then $|a| + |b| = a+b = |a + b| \le |a+b| + |a-b|$.

If $a \ge 0$ and $b < 0$ then $|a| + |b| = a - b = |a-b| \le |a + b| + |a-b|$.

If $a < 0$ and $b \ge 0$ then $|a| + |b| = -a + b = |b-a| = |a-b| \le |a + b| + |a-b|$.

If $a< 0$ and $b < 0$ then $|a| + |b| = -a - b = |-a -b | = |a+b| \le |a+b| + |a-b|$.


We have: $(|a| + |b|)^2 = a^2 + 2|ab| + b^2$, and $(|a+b|+|a-b|)^2 = (a+b)^2 + 2|a^2-b^2| + (a-b)^2 = 2(a^2+b^2) + 2|a^2-b^2|$. Thus we prove: $|ab| - |a^2-b^2| \leq \dfrac{a^2 +b^2}{2}$. We have: $|ab| - |a^2-b^2| \leq |ab| \leq \dfrac{a^2+b^2}{2}$, and the inequality is proved.

  • $\begingroup$ Where does the $|ab|\leq \frac{a^2+b^2}{2}$ come from? $\endgroup$ – George Oct 22 '14 at 16:58
  • $\begingroup$ I think you've lost a factor of two halfway through: $2|ab|$ somehow became $|ab|$. $\endgroup$ – Mark Dickinson Oct 22 '14 at 17:04
  • $\begingroup$ That does not use triangular inequality and that is indeed what the Op precisely did already. $\endgroup$ – Guy Fsone Nov 19 '17 at 6:00

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