Jtdcftul

Suppose that $ab leq frac{1}{p}a^p+frac{1}{q}b^q$ holds for every real numbers $a,bge 0$. (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.

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.

I think the following can help.

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

I found an argument that works for $p,qge 1$, since I need convexity of $xto x^p,x^q$.
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
$$
a^{p'}/p'=maxlimits_{b>0}{ab-b^q/q}
$$
(just because $ato a^{p'}/p'$ is the conjugate (Legendre transform) of $bto b^q/q$). Then we can pick $a$ large enough such that
$$
frac{a^p}{p}<frac{a^{p'}}{p'}
$$
and for such $a$ we would have
$$
a^{p}/p<maxlimits_{b>0}{ab-b^q/q}
$$
which implies that for some $b>0$ (and the given $a$) Young's inequality is violated.

Hope this helps.