Norm on unitisation is submultiplicative Let $A$ be a $C^\ast$-algebra and let $\widetilde{A}$ denote its unitisation. Define a norm on $\widetilde{A}$ as $\|(a,\lambda)\| = \sup_{\|b\|=1} \|ab + \lambda b\|$. 
I could show that $\|(a,\lambda)^\ast (a,\lambda)\| = \|(a,\lambda)\|^2$ using the submultiplicativity of the norm but now I'm having trouble showing it is submultiplicative. 
Please could someone help me and show me how to show 
$\displaystyle \|(a,\lambda)(b,\mu)\| \le \|(a,\lambda)\| \|(b,\mu)\|$?

What you are doing here is that you identify the element $(a, \lambda)$ with the linear operator $T_{(a,\lambda)} : x\mapsto ax + \lambda x$.
The norm $\Vert (a,\lambda) \Vert$ is then the operator norm $\Vert T_{(a,\lambda)} \Vert$.
It is easy to see that the operator norm in general is submultiplicative.
First show that $\Vert Tx \Vert \leq \Vert T\Vert \cdot \Vert x \Vert$ for all $x$ (using homogeneity of the norm).
This implies $\Vert TS x \Vert \leq \Vert T\Vert \Vert Sx\Vert \leq \Vert T\Vert \cdot \Vert S\Vert \cdot \Vert x \Vert$, which yields the submultiplicativity.
