Young's convolution inequality states that if $f\in L^p(X,\mu), g\in L^q$, and $$ 1/p+1/q=1/r+1, $$ then $\|f\|_p\|g\|_q\geq\|f\ast g\|_r.$
Now, if we write $\frac{1}{r}=\frac{\theta}{p}+\frac{1-\theta}{q},$ then it is easy to deduce that $$ \frac{1-\theta}{p}+\frac{\theta}{q}=1. $$ This relation somehow tells us that $1/r$ is the "reflection" of the number $1$ in the midpoint of $1/p$ and $1/q$, because we are swapping $1-\theta$ and $\theta$.
Is there an intuitive, possibly geometric explanation of this observation? Why the inequality works this way?
I'm not sure if this is sufficiently intuitive or geometric, but it might be helpful for thinking about the play of exponents in Young's inequality.
Now the exponent condition can equivalently be written as $1/p' + 1/q' = 1/r'$, where the primes indicate dual exponents. Here we can draw in the Hausdorff-Young inequality, which gives that the Fourier transform maps $L^s$ into $L^{s'}$ for $1 \leq s \leq 2$.
So at least if $1 \leq p, q \leq 2$, then $f \in L^p$ and $g \in L^q$ implies that $\hat{f} \in L^{p'}$ and $\hat{g} \in L^{q'}$; and of course the Fourier transform converts convolution to multiplication, so $(f \ast g)^{\wedge} = \hat{f} \hat{g}$. By Hölder's inequality, we can then see that $(f \ast g)^{\wedge} \in L^{r'}$, which is at least consistent with $f \ast g \in L^r$.