Theorem: Young's Inequality for Products

Let $p, q \in \R_{> 0}$ be strictly positive real numbers such that:
:$\dfrac 1 p + \dfrac 1 q = 1$
Then:
:$\forall a, b \in \R_{\ge 0}: a b \le \dfrac {a^p} p + \dfrac {b^q} q$
Equality occurs {{iff}}:
:$b = a^{p - 1}$

Proof:
:400px
In the above diagram:
:the {{Color|blue}} region corresponds to $\ds \int_0^\alpha t^{p - 1} \rd t$
:the {{Color|red}} region corresponds to $\ds \int_0^\beta u^{q - 1} \rd u$.
From Positive Real Numbers whose Reciprocals Sum to 1, it is necessary for both $p > 1$ and $q > 1$.
{{begin-eqn}}
{{eqn | l = \frac 1 p + \frac 1 q
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = p + q
      | r = p q
      | c = multiplying both sides by $p q$
}}
{{eqn | ll= \leadsto
      | l = p + q - p - q + 1
      | r = p q - p - q + 1
      | c = adding $1 - p - q$ to both sides
}}
{{eqn | ll= \leadsto
      | l = 1
      | r = \paren {p - 1} \paren {q - 1}
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | l = \frac 1 {p - 1}
      | r = q - 1
      | c = 
}}
{{end-eqn}}
Accordingly:
:$u = t^{p - 1} \iff t = u^{q - 1}$
Let $a, b$ be any positive real numbers.
Since $a b$ is the area of the rectangle in the given figure, we have:
:$\ds a b \le \int_0^a t^{p - 1} \rd t + \int_0^b u^{q - 1} \rd u = \frac {a^p} p + \frac {b^q} q$
Note that even if the graph intersected the side of the rectangle corresponding to $t = a$, this inequality would still hold.
Also note that if either $a = 0$ or $b = 0$ then this inequality holds trivially.
{{qed|lemma}}
It remains to prove the equality condition:
:$a b = \dfrac {a^p} p + \dfrac {b^q} q$
{{iff}}:
:$b = a^{p - 1}$
=== Necessary Condition ===
Let $b = a^{p - 1}$.
Then:
{{begin-eqn}}
{{eqn | l = \frac {a^p} p + \frac {b^q} q
      | r = \frac {a^p} p + \frac {\paren {a^{p - 1} }^q} q
      | c = {{hypothesis}}
}}
{{eqn | r = \frac {a^p} p + \frac {\paren {a^{p/q} }^q} q
      | c = {{hypothesis}}: $\dfrac 1 p + \dfrac 1 q = 1$ 
}}
{{eqn | r = a^p \paren {\frac 1 p + \frac 1 q}
      | c = simplifying
}}
{{eqn | r = a \cdot a^{p - 1}
      | c = {{hypothesis}}: $\dfrac 1 p + \dfrac 1 q = 1$
}}
{{eqn | r = a b
      | c = {{hypothesis}}
}}
{{end-eqn}}
That is:
:$b = a^{p - 1} \implies a b = \dfrac {a^p} p + \dfrac {b^q} q$
{{qed|lemma}}
=== Sufficient Condition ===
Let:
:$a b = \dfrac {a^p} p + \dfrac {b^q} q$
{{AimForCont}} $b \ne a^{p - 1}$.
==== Case $1$: $b < a^{p - 1}$ ====
Referring to the diagram above, let us identify:
:$a = \alpha$
:$b = \beta$
Thus $b < a^{p - 1}$ is precisely what is diagrammed in the image above.
We note that the sum of the {{Color|blue}} and {{Color|red}} regions strictly exceeds the area of the rectangle contained by $\alpha$ and $\beta$.
Hence we have that:
:$a b < \dfrac {a^p} p + \dfrac {b^q} q$
==== Case $2$: $b > a^{p - 1}$ ====
We draw a diagram similar to the one above, although integrated {{WRT|Integration}} $u$ rather than $t$.   
This shows that:
:$b \ne a^{p - 1} \implies a b \ne \dfrac {a^p} p + \dfrac {b^q} q$
These two cases together contradict our assertion that $a b = \dfrac {a^p} p + \dfrac {b^q} q$.
Hence by Proof by Contradiction:
:$a b = \dfrac {a^p} p + \dfrac {b^q} q \implies b = a^{p - 1}$
{{qed}}