Why the relation between the exponents in Young's inequality is inevitable? Suppose someone told you that there exist a pair of positive real numbers $p,q$ such that for every positive real numbers $a,b$, the following inequality holds
$$ ab \leqslant \frac{a^p}{p} + \frac{b^q}{q}. $$
Would you have an immediate way of deducing that $\frac{1}{p}+\frac{1}{q}=1$?
i.e. I am looking for elegant ways to see why the relation between the exponents in Young's inequality is forced/inevitable. (hopefully with as few computations as possible).
Here is my way of seeing this:
Taking $a^p=b^q$, we get
$$ a^{\frac{p+q}{q}}=aa^{\frac{p}{q}}=ab \le a^p(\frac{1}{p}+\frac{1}{q}),$$
which is equivalent to $$ a^{r} \le \frac{1}{p}+\frac{1}{q},$$
where $r=\frac{p+q-pq}{q}$. Now, if $r \neq 0$, then taking $a \to 0$ or $a \to \infty$, depending on the sign of $r$,we get a contradiction. This shows that $r$ must be zero, and we are done.
Is there a simpler way?
Comment: By plugging in $a=b=1$, we immediately see that $1 \le \frac{1}{p}+\frac{1}{q}$.

One way is to see Young's inequality as describing an inequality between areas.  Consider the graph of $y=x^{p-1}$, in the domain $x \in [0, a]$.  Let the area under this curve be $A$.  Now as $x^{p-1}$ is increasing for $p> 1$, we may consider the area to the left of the curve, say for $y \in [0, b]$, and let this be $B$. 
Young's inequality is equivalent to saying $A+B \geqslant ab$, the area of the full rectangle, and equality is iff $b = a^{p-1}$.    Work out what $B$ is and you will see how $q$ and the condition $\frac1p+\frac1q=1$ emerges.  For an illustration, the case $a=3, b=6, p = 3$ is drawn below.

Note the shaded green part represents the excess in the inequality.
