Let $A$ be a $C^*$-algebra. Consider the unitalization $A^+$ with underlying vector space $A^+=A\oplus \mathbb{C}$. The norm on $A^+$ is defined as $||a+\lambda||_{A^+}=\mathrm{sup}_{||b||\leq 1}\ ||ab+\lambda b||_A$. In order to show that this is positive definite I consider $(a,\lambda)\in A^+$ such that $||a+\lambda||_{A^+}=0$. Without loss of generality $||a||\leq 1$. This means that for all $b\in A$ with $||b||\leq 1$ we have $ab+\lambda b=0$ using positive definiteness of the norm on $A$. Now $b=a$ implies $a^2=-\lambda a$ and therefore $|\lambda|=||a||$. Therefore it suffices to show that $a$ or $\lambda$ equals zero.
Can anybody help me with the rest? In every book or paper a read, there is no proof of this.
If $\lambda=0$, take $b=a^*$ and you get $aa^*=0$ and hence $a=0$.
If $\lambda\ne0$, then $ab+\lambda b$ can be read as $cb=b$, where $c=-a/\lambda$. Taking adjoints it can also be seen as $bd=d$, where $d=-a^*/\overline\lambda$. This implies that $-a/\lambda=1$, so $a=-\lambda$. This is in contradiction by construction of the unitization, where the only element of $a$ that is a scalar is zero.