(I learned from Terry Tao the following proof, which exploits a symmetry to simplify the task of proving an estimate by normalising one or more inconvenient factors to equal $1$.)
I assume here that $1\leq p<\infty$. We want to show that
$$
\|x+y\|_p\leq\|x\|_p+\|y\|_p\tag{*}
$$
When the RHS is $0$, the proof is trivial. Suppose it is positive. By homogeneity $\|cx\|_p=|c|\|x\|_p$ we may reduce to the case $\|x\|_p=1-\lambda$ and $\|y\|_p=\lambda$ for some $0\leq\lambda\leq 1$. The cases $\lambda=0,1$ are trivial, so suppose $0<\lambda<1$. Writing $X:=x/(1-\lambda)$ and $Y:=y/\lambda$ we reduce to the convexity estimate:
$$
\|(1-\lambda)X+\lambda Y\|_p\leq 1\quad\text{whenever } \|X\|_p=\|Y\|_p=1\
\text{and }0\leq\lambda\leq 1.
$$
But since $z\mapsto|z|^p$ is convex for $p\geq 1$, we have the coordinate-wise convexity bound
$$
|(1-\lambda)X_i+\lambda Y_i|^p\leq (1-\lambda)|X_i|^p+\lambda |Y_i|^p.
$$
Summing $i$ from $1$ to $n$, we obtain
$$
\|(1-\lambda)X+\lambda Y\|_p^p\leq 1
$$
and thus the claim follows.
Note that this proof works for the general abstract $L^p$ spaces as well.
We will assume that $f$ and $g$ are real-valued. The proof for complex-valued $f$ and $g$ follows analogously.
Let $||f||_1^2\equiv \int_a^b|f|^2dx+\int_a^b|f'|^2dx$. Then, we have
$$\begin{align}
||f+g||_1^2&=\int_a^b|f+g|^2\,dx+\int_a^b|f'+g'|^2\,dx\\\\
&=\int_a^b(|f|^2+|f'|^2)\,dx+\int_a^b (|g|^2+|g'|^2)\,dx \\\\
&+2\int_a^b(fg+f'g')\,dx \tag 1\\\\
&\le \int_a^b(|f|^2+|f'|^2)\,dx+\int_a^b (|g|^2+|g'|^2)\,dx\\\\
&+2\sqrt{\int_a^b(|f|^2+|f'|^2)\,dx}\sqrt{\int_a^b(|g|^2+|g'|^2)dx} \tag 2\\\\
&=\left(\sqrt{\int_a^b(|f|^2+|f'|^2)\,dx}+\sqrt{\int_a^b(|g|^2+|g'|^2)\,dx}\right)^2\\\\
&=\left(||f||_1+||g||_1\right)^2
\end{align}$$
as was to be shown!
NOTE:
In going from $(1)$ to $(2)$, we used the Cauchy-Schwarz Inequality. To prove the inequality explicitly here, we first note that
$$\int_a^b\left(f-\frac{\int_a^b(fg+f'g')\,dx'}{\int_a^b(g^2+g'^2)\,dx'}\,g\right)^2\,dx+\int_a^b\left(f'-\frac{\int_a^b(fg+f'g')\,dx'}{\int_a^b(g^2+g'^2)\,dx'}\,g'\right)^2 \ge 0$$
since the integrands of both integrals are non-negative. Then, completing the squares of both integrands and gathering terms yields
$$\left(\int_a^b(f^2+f'^2)\,dx\right)\left(\int_a^b(g^2+g'^2)\,dx\right)\ge \left(\int_a^b(fg+f'g')\,dx\right)^2 \tag 3$$
Taking the square root of both sides of $(3)$, we obtain the desired result
$$\int_a^b(fg+f'g')\,dx\le \sqrt{\left(\int_a^b(f^2+f'^2)\,dx\right)}\sqrt{\left(\int_a^b(g^2+g'^2)\,dx\right)}$$
Best Answer
Any inner product satisfies the C-S inequality $ |\langle x, y \rangle| \leq \|x\| \|y\|$ from which you can complete your argument immediately.