Suppose that $ab \leq \frac{1}{p}a^p+\frac{1}{q}b^q$ holds for every real numbers $a,b\ge 0$. (where $p,q>0$ are some fixed numbers).
Is it true that $ \frac{1}{p}+\frac{1}{q}=1$?
I guess so, and I would like to find an easy proof of that fact. Plugging in $a=b=1$, we get $ \frac{1}{p}+\frac{1}{q}\ge1$. Is there an easy way to see that the converse inequality must hold?
To be clear, I am not looking for proofs of Young's inequality; only for a way to see why the relation between the conjugate exponents is the only possible option.
Edit:
Here is a comment to help future readers (including future me) to see how can one come with Nicolás's nice idea to apply the inequality for $a=\lambda^{\frac{1}{p}}, b=\lambda^{\frac{1}{q}}$.
The idea is that it is inconvenient to compare a sum with another number; By using the specific choice of $a,b$ above, the sum is simplified, since the scales of the auxiliary parameter $\lambda$ in both summands are now identical.
Comment: I know that the relation $ \frac{1}{p}+\frac{1}{q}=1$ is necessary for Holder's inequality in general; this can be seen by scaling the measure. However, I don't think this approach is applicable here.
Best Answer
If you apply the inequality for $a=\lambda^{\frac{1}{p}}, b=\lambda^{\frac{1}{q}}$, you get something like $$ \lambda^{\frac{1}{p}+\frac{1}{q}}\leq \left( \frac{1}{p}+\frac{1}{q} \right)\lambda, \qquad \forall \lambda>0. $$ If you take $\lambda \to \infty$, it's clear that $\frac{1}{p}+\frac{1}{q}\leq 1$. Otherwise, after dividing by $\lambda$ you get $$ \lambda^{\frac{1}{p}+\frac{1}{q}-1} \leq \frac{1}{p}+\frac{1}{q}, $$ which is false for $\lambda$ large enough, as the right side is constant. Taking $\lambda \to 0$ gets the other inequality.