Proving that $|a-b|≤|a|+|b|$

Question

Can someone prove this to me: $$|a-b|≤|a|+|b|$$ I am in 8th grade and I have this for my homework. Thanks for helping.

Answer 1

Square both members: you can because the numbers involved are non negative. You get the equivalent inequality $$ a^2-2ab+b^2\le a^2+2|ab|+b^2 $$ (justify the passages), that reduces to the equivalent inequality $$ -ab\le |ab| $$

Is this inequality true?

Answer 2

HINT: There are much slicker, shorter arguments, but the most straightforward approach is to divide it into cases:

• $a\le b\le 0$;
• $a\le 0\le b$;
• $0\le a\le b$;
• $b\le a\le 0$;
• $b\le 0\le a$; and
• $0\le b\le a$.

Check that these cover all possibilities, and note that in each case you can get rid of the absolute values.

Answer 3

Hint: $a-b=a+(-b)$. I suppose you know the triangle inequality.

If you have to prove it from scratch, another hint:

$\lvert x\rvert \le\lvert y\rvert \iff\lvert x\rvert^2 \le\lvert y\rvert^2\iff x^2\le y^2$.