I'm having some trouble seeing how two inequalites have been combined in a proof i'm going through. The first inequlaity is
$$\| w - w_h \|_a \le \|\phi\|_{L^2}$$
and the second inequality is given by
$$\|w\|_a \le C \|\phi\|_{L^2} .$$
Which combine to give
$$\|w_h\|_a \le (C + 1) \|\phi\|_{L^2}$$
Where $w \in H^1_0$ is the solution to a PDE and $w_h$ is the finite element solution. The function $\phi \in L^2 $ has the standard $L^2$ norm and the $a$ norm is given by $\|w\|_a = \sqrt{a(w,w)}$
Any help will be appreciated.
Edit: My Attempt
I'm not sure if this is correct,
\begin{align} \|w_h\|_a &= \|(-(w - w_h))+w \|_a \\ &\le \| -(w-w_h)\|_a + \|w\|_a \\ &= \| w-w_h\|_a + \|w\|_a \\ &\le \|\phi\|_{L^2} + C\|\phi\|_{L^2} \end{align}
where i used the fact that for any norm $\|\alpha f\| = |\alpha| \|f\|$ for $\alpha \in \mathbb{R}$.
By the triangle inequality,
\begin{align} \|w_h\|_a &= \|w_h-w+w\|_a \\ &\le \| w-w_h\|_a + \|w\|_a \\ &\le \|\phi\|_{L^2} + C\|\phi\|_{L^2} \end{align}