In proving that the square-integrable functions $\varphi: \mathbb{R} \to \mathbb{C}$ can be made into a complex vector space, my book endeavors to prove that the space is closed under superposition. Thus, it writes $$\int_{-\infty}^{\infty} |\varphi(x) + \alpha \psi(x)|^2 \, dx \leq \int_{-\infty}^{\infty} |\varphi(x)|^2 \, dx + |\alpha^2|\int_{-\infty}^{\infty} |\psi(x)|^2 \, dx < \infty, $$ but I can't for the life of me understand what is going on in this second step. I imagined that we should have used the triangle inequality, $$|\varphi(x) + \alpha \psi(x)|^2 \leq |\varphi(x)|^2 + |\alpha|^2| \psi(x)|^2 + 2|\varphi(x)||\alpha|| \psi(x)|,$$ but of course this last term on the RHS does not appear above.
What is going on? Is there some basic complex analysis that I don't know which is being used here?
The inequality, as written, is wrong (just check for $\psi=\varphi$). I think the author is just confused because the map $\lVert f\rVert_2= \sqrt{\int_{-\infty}^\infty \lvert f(x)\rvert^2\,dx}$ is subadditive. But here we are considering its square!
What should be done is, as you suggest, $$\lVert \varphi+\alpha\psi\rVert_2^2\le \lVert \varphi\rVert_2^2+\lvert\alpha\rvert^2\lVert\psi \rVert_2^2+2\lvert\alpha\rvert\int_{-\infty}^\infty \lvert \varphi(x)\psi(x)\rvert\,dx\\\le\lVert \varphi\rVert_2^2+\lvert\alpha\rvert^2\lVert\psi \rVert_2^2 +2\alpha(\lVert\psi \rVert_2\lVert\varphi \rVert_2)$$
The last inequality being Cauchy-Schwarz.