Does Young's inequality hold only for conjugate exponents?

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