One proof of the Schwarz inequality on $\mathbb{R}^n$ is to note that $$(\sum x_i^2)(\sum y_i^2) = (\sum x_i y_i)^2 + \sum_{i<j}(x_iy_j - x_jy_i)^2.$$ Spivak's Calculus, 4th ed., exercise 13-39(b), asks us to adapt this discrete proof to a proof that $$\left( \int_a^b fg \right)^2 \leq \left( \int_a^b f^2 \right) \left( \int_a^b g^2 \right)$$ for $f,g$ Riemann integrable functions on $[a,b]$. Here is my attempt; I wonder if this is the way you would do it:
First, I claim a lemma:
Lemma: To show that $\left( \int_a^b fg \right)^2 \leq \left( \int_a^b f^2 \right) \left( \int_a^b g^2 \right)$, it is sufficient to show that $\left( \mathcal{R}(fg,P)\right)^2 \leq \mathcal{R}(f^2, P)\mathcal{R}(g^2,P)$ for any tagged partition $P$, where $\mathcal{R}$ is the Riemann sum.
Now note that $$\sum A_i^2 (t_i - t_{i-1})\sum B_i^2 (t_i - t_{i-1}) = \sum \left( A_iB_i (t_i -t_{i-1})\right)^2 \\ + \sum_{i<j}(t_i - t_{i-1})(t_j - t_{j-1})(A_iB_j-A_jB_i)^2.$$ Let $A_i =f(x_i)$ and $B_i = g(x_i)$, and indeed we see that the inequality holds for all tagged partitions $P$.
Is this how you would do it? Another question: it would be more elegant if I were able to adapt this proof to show that if $f \neq \lambda g$, and $f,g$ are continuous, then the inequality is strict. But all the ways I'm thinking of to show this are getting hopelessly complicated. Does anyone have a suggestion? (I can think of other ways to show that the inequality is strict, but I'd like to adapt this proof to do so.)
Addendum: @ThisIsMuchHealthier provided an excellent answer below, which relies on the pointwise equality $$f(x)^2g(y)^2 + f(y)^2g(x)^2 = 2f(x)g(x)f(y)g(y) + (f(x)g(y)-f(y)g(x))^2.$$
How can we adapt this to a proof for $f,g:\mathbb{R}\to \mathbb{C}$? The analogous equality would read $$|f(x)|^2|g(y)|^2 + |f(y)|^2|g(x)|^2 = 2 \Re \left( f(x)\overline{g(x)}\overline{f(y)}g(y) \right) + |f(x)g(y)-f(y)g(x)|^2.$$ We'd like to turn $2 \Re \left( f(x)\overline{g(x)}\overline{f(y)}g(y) \right) $ into $2f(x)\overline{g(x)}\overline{f(y)}g(y)$ somehow, but it's not clear how to do that. There is a discrete version of this proof for $\mathbb{C}^n$ that relies on $$(|x_1|^2 + \dotsb +|x_n|^2)(|y_1|^2 + \dotsb + |y_n|^2) = |x_1\overline{y_1} + \dotsb + x_n\overline{y_n}|^2 + \frac12 \sum_{i, j} |x_iy_j - x_jy_i|^2,$$ which suggests that there should be a way to adapt this for the integral. Any ideas?
Edit: I think the answer is that $2 \Re \left( f(x)\overline{g(x)}\overline{f(y)}g(y) \right)$ is sufficient, because $$\int_a^b\int_a^b 2 \Re \left( f(x)\overline{g(x)}\overline{f(y)}g(y) \right) dxdy= 2 \Re \int_a^b\int_a^b f(x)\overline{g(x)}\overline{f(y)}g(y)dxdy \\= 2 \int_a^b\int_a^b f(x)\overline{g(x)}\overline{f(y)}g(y)dxdy,$$ because $$\int_a^b\int_a^b f(x)\overline{g(x)}\overline{f(y)}g(y)dxdy = \int_a^b f(x)\overline{g(x)}dx \int_a^b \overline{f(y)}g(y)dy\\ = \left( \int_a^b f(x)\overline{g(x)}dx\right) \overline{\left( \int_a^b f(y)\overline{g(y)}dy \right)}\\=\left| \left( \int_a^b f(x)\overline{g(x)}dx\right) \right|^2$$ is real.
Addendum 2: The user 900 Sit-ups a day (formerly "This is much healthier") challenged me to come up with a similar identity for vector-valued functions $\mathbf{f},\mathbf{g}$.
We have the identity $$(f_i(t)g_j(s)-f_j(s)g_i(t))^2 = f_i(t)^2g_j(s)^2 + f_j(s)^2g_i(t)^2 - 2f_i(t)g_i(t)f_j(s)g_j(s).$$ Summing over all $i,j$ gives $$|\mathbf{f}(t)|^2|\mathbf{g}(s)|^2+|\mathbf{f}(s)|^2|\mathbf{g}(t)|^2=2\langle \mathbf{f}(t), \mathbf{g}(t) \rangle \langle \mathbf{f}(s),\mathbf{g}(s) \rangle + \sum_{i,j}(f_i(t)g_j(s) - f_j(s)g_i(t))^2 \\ \int_a^b \int_a^b |\mathbf{f}(t)|^2|\mathbf{g}(s)|^2+|\mathbf{f}(s)|^2|\mathbf{g}(t)|^2 dsdt= \int_a^b \int_a^b 2\langle \mathbf{f}(t), \mathbf{g}(t) \rangle \langle \mathbf{f}(s),\mathbf{g}(s) \rangle \\+ \sum_{i,j}(f_i(t)g_j(s) - f_j(s)g_i(t))^2ds dt \\ 2\int_a^b |\mathbf{f}|^2 \int_a^b |\mathbf{g}|^2 = 2 \left( \int_a^b \langle \mathbf{f},\mathbf{g} \rangle\right)^2 + \int_a^b \int_a^b \sum_{i,j}\left(f_i(t)g_j(s) - f_j(s)g_i(t) \right)^2ds dt\\ \int_a^b |\mathbf{f}|^2 \int_a^b |\mathbf{g}|^2 = \left( \int_a^b \langle \mathbf{f},\mathbf{g} \rangle\right)^2 + \frac12 \int_a^b \int_a^b \sum_{i,j}\left(f_i(t)g_j(s) - f_j(s)g_i(t) \right)^2ds dt$$
I don't like $i<j$ in the double sum; better to divide by $2$ instead. The integral version of this identity is $$\begin{split}\frac12 \iint f^2(x)g^2(y)+f^2(y)g^2(x)\,dx\,dy &= \iint f(x)g(x)f(y)g(y)\,dx\,dy \\ &+ \frac12 \iint (f(x)g(y)-f(y)g(x))^2\,dx\,dy \end{split}$$ The equality here holds pointwise, i.e., it is an algebraic identity with integral signs in front.
By Fubini's theorem, the left hand side is $\left(\int f^2(x) \,dx\right)\left(\int g^2(x) \,dx\right)$ while the first integral on the right is $\left(\int f(x)g(x)\,dx\right)^2$.
Equality holds iff $f(x)g(y)=f(y)g(x)$ for almost all $(x,y)$. Fixing $y$ and varying $x$, conclude that $\lambda f+\mu g=0$ a.e. for some constants $\lambda,\mu$.