Young's Inequality for Products/Geometric Proof

From ProofWiki
Jump to navigation Jump to search

Theorem

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 if and only if:

$b = a^{p - 1}$

Proof

Holder's Ineq.jpg

In the above diagram:

the $\color { blue } {\text {blue} }$ region corresponds to $\ds \int_0^\alpha t^{p - 1} \rd t$
the $\color { red } {\text {red} }$ 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 $q > 1$.

\(\ds \frac 1 p + \frac 1 q\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds p + q\) \(=\) \(\ds p q\) multiplying both sides by $p q$
\(\ds \leadsto \ \ \) \(\ds p + q - p - q + 1\) \(=\) \(\ds p q - p - q + 1\) adding $1 - p - q$ to both sides
\(\ds \leadsto \ \ \) \(\ds 1\) \(=\) \(\ds \paren {p - 1} \paren {q - 1}\) simplifying
\(\ds \leadsto \ \ \) \(\ds \frac 1 {p - 1}\) \(=\) \(\ds q - 1\)

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.

$\Box$


It remains to prove the equality condition:

$a b = \dfrac {a^p} p + \dfrac {b^q} q$

if and only if:

$b = a^{p - 1}$


Necessary Condition

Let $b = a^{p - 1}$.

Then:

\(\ds \frac {a^p} p + \frac {b^q} q\) \(=\) \(\ds \frac {a^p} p + \frac {\paren {a^{p - 1} }^q} q\) by hypothesis
\(\ds \) \(=\) \(\ds \frac {a^p} p + \frac {\paren {a^{p/q} }^q} q\) by hypothesis: $\dfrac 1 p + \dfrac 1 q = 1$
\(\ds \) \(=\) \(\ds a^p \paren {\frac 1 p + \frac 1 q}\) simplifying
\(\ds \) \(=\) \(\ds a \cdot a^{p - 1}\) by hypothesis: $\dfrac 1 p + \dfrac 1 q = 1$
\(\ds \) \(=\) \(\ds a b\) by hypothesis

That is:

$b = a^{p - 1} \implies a b = \dfrac {a^p} p + \dfrac {b^q} q$

$\Box$


Sufficient Condition

Let:

$a b = \dfrac {a^p} p + \dfrac {b^q} q$

Aiming for a contradiction, suppose $b \ne a^{p - 1}$.


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 $\color { blue } {\text {blue} }$ and $\color { red } {\text {red} }$ 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 with respect to $u$ 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:

$a b = \dfrac {a^p} p + \dfrac {b^q} q \implies b = a^{p - 1}$

$\blacksquare$


Source of Name

This entry was named for William Henry Young.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Young%27s_Inequality_for_Products/Geometric_Proof&oldid=658517"