Proof of Young's Inequality
Theorem
Given constants $\displaystyle {{1} \over {p}} + {{1} \over {q}} = 1$ that satisfy the condition and two positive numbers $p,q$ and $a,b$ which are greater than 1,
$$ ab \le { {a^{p}} \over {p} } + {{b^{q}} \over {q}} $$
Description
Apart from the aesthetically pleasing algebraic aspect, this inequality is not often mentioned except when proving the Hölder inequality.
Proof
Since both $a$ and $b$ are positive, there exists a real number $A,B$ that satisfies $a = e^A, b = e^B$.
Convexity and the Second Derivative
Let’s say that $f$ is twice differentiable at $I$. For $f$ to be convex at $I$, $f '' (x) >0$ is a necessary and sufficient condition.
Meanwhile, since $ e^x>0$, the second derivative is always positive, therefore, it is convex at $\mathbb{R}$.
For a convex $f : I \to \mathbb{R}$ at $I \subset \mathbb{R}$ and for $\displaystyle \sum_{k=1}^{n} \lambda_{k} = 1, \lambda_{k}>0$, $f( \lambda_{1} x_{1} + \lambda_{2} x_{2} + \cdots + \lambda_{n} x_{n} ) \le \lambda_{1} f( x_{1}) + \lambda_{2} f( x_{2}) + \cdots + \lambda_{n} f( x_{n} )$.
By Jensen’s inequality,
$$ e^{{{1} \over {p}} p A + {{1}\over {q}} q B} \le {{1} \over {p}} e^{pA} + {{1} \over {q}} e^{qB} $$
In summary, $$ e^{A+B} \le {{1} \over {p}} (e^{A})^p +{{1} \over {q}} (e^{B})^q $$
Since it was $a = e^A, b = e^B$,
$$ ab \le { {a^{p}} \over {p} } + {{b^{q}} \over {q}} $$
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