I am reading C*-algebras and Operator Theory by Gerard Murphy. It defines an algebra as a vector space $A$ together with a bilinear map $$A^2\to A,\;\;\;(a,b)\mapsto ab,$$ such that $$a(bc)=(ab)c\;\;\;(a,b,c\in A).$$ If a norm $\|\cdot\|$ on $A$ is submultiplicative, i.e. that $$\|ab\|\leq\|a\|\|b\|\;\;\;(a,b\in A),$$ the pair $(A,\|\cdot\|)$ is called a normed algebra.
Murphy then uses the inequality $$\|ab-a'b'\|\leq\|a\|\|b-b'\|+\|a-a'\|\|b'\|$$ to prove that the multiplication $(a,b)\mapsto ab$ is jointly continuous. How does this inequality come about? Had there been a distributive law, one could have done like this: $$\|ab-a'b'\|\leq\|ab-ab'\|+\|ab'-a'b'\|=\|a(b-b')\|+\|(a-a')b'\|\\\leq\|a\|\|b-b'\|+\|a-a'\|\|b'\|.$$ But, at least a priori, there isn't. So how should one attack this?