Does Young's inequality hold only for conjugate exponents?

Question

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.

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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.

Answer 1

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.

Answer 2

I found an argument that works for $p,q\ge 1$, since I need convexity of $x\to x^p,x^q$. Assume by contradiction that $1/p+1/q>1$. Then, if $p'$ is the conjugate of $q$ we know that $p'>p$.

Now, for any $a>0$ we have $$ a^{p'}/p'=\max\limits_{b>0}\{ab-b^q/q\} $$ (just because $a\to a^{p'}/p'$ is the conjugate (Legendre transform) of $b\to b^q/q$). Then we can pick $a$ large enough such that $$ \frac{a^p}{p}<\frac{a^{p'}}{p'} $$ and for such $a$ we would have $$ a^{p}/p<\max\limits_{b>0}\{ab-b^q/q\} $$ which implies that for some $b>0$ (and the given $a$) Young's inequality is violated.

Hope this helps.

Answer 3

I think the following can help.

By AM-GM $$\frac{1}{p}a^p+\frac{1}{q}b^q\geq\left(\frac{1}{p}+\frac{1}{q}\right)\left(\left(a^p\right)^{\frac{1}{p}}\left(b^q\right)^{\frac{1}{q}}\right)^{\frac{1}{\frac{1}{p}+\frac{1}{q}}}=\left(\frac{1}{p}+\frac{1}{q}\right)(ab)^{\frac{1}{\frac{1}{p}+\frac{1}{q}}}.$$ The equality occurs, of course.