The Cauchy Schwarz inequality states that:
$$
\left|\int_{-\infty}^{\infty}g_1(t)g_2(t) dt\right|^2 \leq \int_{-\infty}^{\infty}|g_1(t)|^2 dt \int_{-\infty}^{\infty}|g_2(t)|^2 dt
$$
I'm going to assume that $f(t)$ is real, just to make the math a little easier. From the above we can write:
$$
\left|\int_{-\infty}^{\infty}f(t)f(t-\tau) dt\right|^2 \leq \int_{-\infty}^{\infty}|f(t)|^2 dt \int_{-\infty}^{\infty}|f(t-\tau)|^2 dt
$$
For the second integral on the right, just use a variable substitution - let $u=t-\tau$ and then rewrite it with the dummy variable $t$ again - we have
$$
\left|\int_{-\infty}^{\infty}f(t)f(t-\tau) dt\right|^2 \leq \int_{-\infty}^{\infty}|f(t)|^2 dt \int_{-\infty}^{\infty}|f(t)|^2 dt
$$
but the integral on the right is just the autocorrelation function i.e. $ \int_{-\infty}^{\infty}|f(t)|^2 dt= R_{f}(0)$. So we can write
$$
\left|\int_{-\infty}^{\infty}f(t)f(t-\tau) dt\right|^2 \leq R^2_{f}(0)
$$
Note that from the definition of $R_f(0)$ it must be a positive quantity, so taking the square root of both sides and paying attention to the positive and negative quantities on the left hand side, we can write
$$
\int_{-\infty}^{\infty}f(t)f(t-\tau) dt \leq R_{f}(0)
$$
Note - you don't have to use the Cauchy Schwarz inequality - another proof:
$$[f(t)-f(t-\tau)]^2 = f^2(t)+f^2(t-\tau) -2f(t)f(t-\tau)
$$
Integrating both sides from $-\infty$ to $\infty$ and using the same trick (variable substitution for the integral over $f^2(t-\tau)$) we can write
$$
\int_{-\infty}^{\infty} [f(t)-f(t-\tau)]^2dt= 2\int_{-\infty}^{\infty}f^2(t)dt-2\int_{-\infty}^{\infty}f(t)f(t-\tau)dt
$$
Rearranging we can write:
$$
\int_{-\infty}^{\infty}f(t)f(t-\tau)dt = \int_{-\infty}^{\infty}f^2(t)dt -\frac{1}{2}\int_{-\infty}^{\infty} [f(t)-f(t-\tau)]^2dt
$$
Now - the second integral on the right has to be a positive quantity and it is subtracted so we have
$$
\int_{-\infty}^{\infty}f(t)f(t-\tau)dt \leq \int_{-\infty}^{\infty}f^2(t)dt
$$
