Proof of Young's Inequality
Theorem
For two constants $p,q$ greater than $1$ satisfying $\displaystyle {{1} \over {p}} + {{1} \over {q}} = 1$ and two positive numbers $a,b$,
$$ ab \le { {a^{p}} \over {p} } + {{b^{q}} \over {q}} $$
Explanation
Apart from its algebraically beautiful form, this is an inequality that is rarely mentioned outside of proving Hölder’s inequality.
Proof
Since both $a$ and $b$ are positive, there exist real numbers $A,B$ satisfying $a = e^A, b = e^B$.
The second derivative of a convex function
Suppose $f$ is twice differentiable on $I$. Then $f$ being convex on $I$ and $f '' (x) >0$ are necessary and sufficient conditions for each other.
Meanwhile, since $ e^x>0$, the second derivative is also always positive, and therefore it is convex on $\mathbb{R}$.
For a convex $f : I \to \mathbb{R}$ on $I \subset \mathbb{R}$ and $\displaystyle \sum_{k=1}^{n} \lambda_{k} = 1, \lambda_{k}>0$, we have $f( \lambda_{1} x_{1} + \lambda_{2} x_{2} + \cdots + \lambda_{n} x_{n} ) \le \lambda_{1} f( x_{1}) + \lambda_{2} f( x_{2}) + \cdots + \lambda_{n} f( x_{n} )$
By Jensen’s inequality,
$$ e^{{{1} \over {p}} p A + {{1}\over {q}} q B} \le {{1} \over {p}} e^{pA} + {{1} \over {q}} e^{qB} $$
Rearranging, $$ e^{A+B} \le {{1} \over {p}} (e^{A})^p +{{1} \over {q}} (e^{B})^q $$
Since $a = e^A, b = e^B$,
$$ ab \le { {a^{p}} \over {p} } + {{b^{q}} \over {q}} $$
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