My question has two parts.
Let $p$ and $q$ be numbers with $p > 1$ and $q > 1$
1) Show $${x^p\over p} + {y^q\over q} \ge {1\over p} + {1\over q}$$ where $x \gt 0$, $y \gt 0$ and $xy = 1$.
2) Use this to verify the following inequality: $$ab \le {a^p\over p} + {b^q\over q}$$ where $a \ge 0$, $b \ge 0$, $p \gt 1$, $q \gt 1$, and ${1\over p} + {1\over q} = 1$.
I can solve the first part using Lagrange multipliers. You eventually get down to $$\begin{align} \require{cancel} &x^p = y^q \\ &x^p = x^{-q}\\ &x^{p+q} = 1\\ \implies &x=1\text{ or }\cancel{p+q=0}\\ \implies &y = 1 \end{align}$$ This suggests that the minimum of ${x^p\over p} + {y^q\over q} $ occurs when $x=y=1$, which is exactly ${1\over p} + {1\over q}$.
I'm not entirely sure how to apply this to the second part. I tried to solve the second part by using Lagrange multipliers (ignoring the conclusion from the first part), but it gets very messy very fast because the variables in the constraint function are $p$ and $q$, not $a$ and $b$.
This makes me think I need to define new variables $x$ and $y$ in terms of $a$ and $b$ or $p$ and $q$ such that $xy=1$ and the equation is similar to the one from part 1. However, I can't find a way to do this that actually seems to help.
Any pointers in the right direction would be appreciated.
Using copper.hat's hint:
$x=\sqrt[p]{a^p\over ab}\\ y=\sqrt[q]{b^q\over ab}$
Show $xy=1$: $$ \begin{align} xy&=\sqrt[p]{a^p\over ab}\sqrt[q]{b^q\over ab}\\ &={a\over (ab)^{{1\over p}}}{b\over (ab)^{{1\over q}}}\\ &={ab\over (ab)^{{1\over p} + {1\over q}}}\\ &={ab\over (ab)^1}\\ &=1 \end{align} $$
From (1) above we now have:
$$ \begin{align} &{x^p\over p} + {y^q\over q} \ge {1\over p} + {1\over q} = 1\\ &{(\sqrt[p]{a^p\over ab})^p\over p} + {(\sqrt[q]{b^q\over ab})^q\over q} \ge 1\\ &{a^p\over abp} + {b^q\over abq} \ge 1\\ &{a^p\over p} + {b^q\over q} \ge ab\\ \end{align} $$