Norm on commutative algebra satisfying square property is submultiplicative

Question

Let $A$ be a commutative algebra and $\| \cdot \|$ be a norm on $A$ such that $\|x^2\|=\|x\|^2$ for all $x \in A$. Show that $\|xy\| \leq \|x\|\|y\|.$

My attempt: If I assume $$2\|xy\|\leq(\|x\|+\|y\|)^2\tag{1}$$ for all $x,y \in A.$

Then from this I get from all $x,y \in A$ $$\|xy\|\leq 2\|x\|\|y\|.$$ Then replacing $x$ and $y$ with $x^{2^n}$ and $y^{2^n}$, the required result follows.

However, I'm not able to prove equation $(1).$ I tried using $4xy=(x+y)^2-(x-y)^2$ but didn't get too far.

Any hints are appreciated.

Answer 1

How did you use $4xy=(x+y)^2-(x-y)^2$? We have \begin{align*} \|4xy\| &=\|(x+y)^2-(x-y)^2\| \\ &\leq\|(x+y)^2\|+\|(x-y)^2\| \\ &=\|x+y\|^2+\|x-y\|^2 \\ &\leq(\|x\|+\|y\|)^2+(\|x\|+\|y\|)^2 \\ &=2(\|x\|+\|y\|)^2, \end{align*} and therefore $$ 2\|xy\| \leq (\|x\|+\|y\|)^2. $$