nyuuzyou/wikis huggingface dataset:
World's Hardest Easy Geometry Problem

== Theorem ==
:
Find $x$.
== Solution ==
We are given that:
:$\angle BAC = 70 \degrees + 10 \degrees = 80 \degrees$
:$\angle ABC = 60 \degrees + 20 \degrees = 80 \degrees$
Thus by Triangle with Two Equal Angles is Isosceles, $\angle ABC$ is isosceles.
That is:
:$AC = BC$
Using Sum of Angles of Triangle equals Two Right Angles we can construct some other angles:
:$\angle ACB = 180 \degrees - 80 \degrees - 80 \degrees = 20 \degrees$
:$\angle BDC = 180 \degrees - 20 \degrees - 20 \degrees = 140 \degrees$
:$\angle AEC = 180 \degrees - 20 \degrees - 10 \degrees = 150 \degrees$
:$\angle AEB = 180 \degrees - 70 \degrees - 80 \degrees = 30 \degrees$
:$\angle ADB = 180 \degrees - 60 \degrees - 80 \degrees = 40 \degrees$
Construct $DF$ parallel to $AB$.
Draw $AF$ crossing $BD$ at $G$.
Draw $CG$.
:
Thus:
:$AD = BF$
:$\angle DAB = \angle FBA$
:$BC$ is common
Hence from Triangle Side-Angle-Side Congruence:
:$\triangle ADB = \triangle AFB$
Immediately:
:$\angle ABG = \angle BAG$
and so:
:$\angle CAG = \angle CBG$
Thus:
:$AC = BC$
:$\angle CAG = \angle CBG$
:$AG = BG$
Hence from Triangle Side-Angle-Side Congruence:
:$\triangle CAG = \triangle CBG$
Thus:
:$\angle ACG = \angle BCG$
But as:
:$\angle ACG + \angle BCG = 20 \degrees$
it follows that $\angle ACG = \angle BCG = 10 \degrees$
See that:
:$\angle ACG = \angle CAE$
:$\angle CAG = \angle ACE$
:$AC$ is common
Hence from Triangle Angle-Side-Angle Congruence:
:$\triangle ACG = \triangle CAE$
and so:
:$(1): \quad CE = AG$
Now we have that:
:$\angle AGB = \angle DGF = 60 \degrees$
:$DG = GF$
So $\triangle DGF$ is equilateral.
Hence:
:$DF = DG = GF$
We have that:
:$\angle ACF = 20 \degrees$
:$\angle CAF = \angle CBD = 20 \degrees$
Hence from Triangle with Two Equal Angles is Isosceles:
:$\triangle ACF$ is isosceles
So:
:$AF = CF$
As $AG = CE$:
:$EF = GF$
and so
From Parallelism implies Equal Corresponding Angles:
:$\angle DFE = \angle ABC = 80 \degrees$
Thus from Sum of Angles of Triangle equals Two Right Angles:
:$\angle FED + \angle FDE = 180 \degrees - \angle DFE = 100 \degrees$
and so:
:$\angle FED + \angle FDE = 50 \degrees$
Then:
:$\angle DEA = \angle FED - \angle AEB$
But $\angle DEA$ is the angle $x$ which was to be found.
Hence the result.
== Also see ==
* Langley's Problem
Category:Triangles
Category:Classic Problems
Woset is Isomorphic to Set of its Initial Segments

== Theorem ==
Let $\struct {S, \preceq}$ be a well-ordered set.
Let:
:$A = \set {a^\prec: a \in S}$
where $a^\prec$ is the strict lower closure of $S$ determined by $a$.
Then:
:$\struct {S, \preceq} \cong \struct {A, \subseteq}$
where $\cong$ denotes order isomorphism.
== Proof ==
Define $f: S \to A$ as:
:$\forall a \in S: \map f a = a^\prec$
where $a^\prec$ is the initial segment determined by $a$.
=== $f$ is Surjective ===
$f$ is trivially surjective by the definition of $A$.
=== $f$ is Strictly Increasing ===
Let $x, y \in S$ with $x \prec y$.
Let $z \in \map f x$.
Then by the definition of initial segment:
:$z \prec x$
By Reflexive Reduction of Ordering is Strict Ordering, $\prec$ is also transitive.
Thus:
:$z \prec y$
Thus by the definition of initial segment:
:$z \in y^\prec = \map f y$
As this holds for all such $z$:
:$\map f x \subseteq \map f y$
As $x \prec y$:
:$x \in y^\prec = \map f y$
But since $\prec$ is antireflexive:
:$x \nprec x$
so:
:$x \notin \map f x$
Thus:
:$\map f x \subsetneqq \map f y$
As this holds for all such $x$ and $y$, $f$ is strictly increasing.
Since a well-ordering is a total ordering, $f$ is an order embedding by Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing.
Thus $f$ is a surjective order embedding and therefore an order isomorphism.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.5$
Category:Well-Orderings
Category:Order Isomorphisms
Wosets are Isomorphic to Each Other or Initial Segments

== Theorem ==
Let $\struct {S, \preceq_S}$ and $\struct {T, \preceq_T}$ be well-ordered sets.
Then precisely one of the following hold:
:$\struct {S, \preceq_S}$ is order isomorphic to $\struct {T, \preceq_T}$
or:
:$\struct {S, \preceq_S}$ is order isomorphic to an initial segment in $\struct {T, \preceq_T}$
or:
:$\struct {T, \preceq_T}$ is order isomorphic to an initial segment in $\struct {S, \preceq_S}$
== Proof Using Choice ==
== Proof Without Using Choice ==
== Also see ==
* Counting Theorem
Category:Well-Orderings
Category:Wosets are Isomorphic to Each Other or Initial Segments
Wosets are Isomorphic to Each Other or Initial Segments/Proof Using Choice

== Theorem ==
== Proof ==
We assume $S \ne \O \ne T$; otherwise the theorem holds vacuously. 
Define:
:$S' = S \cup \text{ initial segments in } S$
:$T' = T \cup \text{ initial segments in } T$
:$\FF = \set {f: S' \to T' \mid f \text{ is an order isomorphism} }$
We note that $\FF$ is non-empty, because it at least contains a trivial order isomorphism between singletons:
:$f_0: \set {\text{smallest element in } S} \to \set {\text{smallest element in } T}$
Such smallest elements are guaranteed to exist by virtue of $S$ and $T$ being well-ordered.
By the definition of initial segment, the initial segments of $S$ are subsets of $S$.
For any initial segment $I_\alpha$ of $S$, such a segment has an upper bound by definition, namely, $\alpha$.
Also, no initial segment of $S'$ is the entirety of $S$.
Thus every initial segment of $S'$ has an upper bound, $S$ itself.
Every chain in $S'$ has an upper bound, because it defines an initial segment. 
The previous reasoning also applies to $T'$.
By Antilexicographic Product of Totally Ordered Sets is Totally Ordered, $S' \times T'$ is itself a totally ordered set, with upper bound $S \times T$.
Thus the hypotheses of Zorn's Lemma are satisfied for $\FF$.
Let $f_1$ be a maximal element of $\FF$.
Call its domain $A$ and its codomain $B$.
Suppose $A$ is an initial segment $I_a$ in $S$ and $B$ is an initial segment $I_b$ in $T$.
Then $f_1$ can be extended by defining $\map {f_1} a = b$.
This would contradict $f_1$ being maximal, so it cannot be the case that both $A$ and $B$ are initial segments.
Then precisely one of the following hold:
:$A = S$, with $S$ order isomorphic to an initial segment in $T$
or:
:$B = T$, with $T$ order isomorphic to an initial segment in $S$
or:
:both $A = S$ and $B = T$, with $S$ order isomorphic to $T$.
The cases are distinct by Well-Ordered Class is not Isomorphic to Initial Segment.
== Sources ==
* $\S \text P.17$
Category:Wosets are Isomorphic to Each Other or Initial Segments
Wronskian of Linearly Dependent Mappings is Zero

== Theorem ==
Let $\map {f_1} x, \map {f_2} x, \dotsc, \map {f_n} x$ be real functions defined on a closed interval $\closedint a b$.
Let $f_1, f_2, \ldots, f_n$ be $n - 1$ times differentiable on $\closedint a b$.
Then:
:$\map W x = 0$
:
:there exists a subinterval of $\closedint a b$ on which $f_1, f_2, \ldots, f_n$ are linearly dependent.
== Proof ==
== Sources ==
* 
* 
Category:Wronskians
Wythoff Pairs are Winning Positions in Wythoff's Game

== Theorem ==
The Wythoff pairs are the winning positions in Wythoff's game.
== Proof ==
== Sources ==
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
Category:Beatty Sequences
X + y + z equals 1 implies xy + yz + zx less than Half

== Theorem ==
Let $x$, $y$ and $z$ be real numbers such that: 
:$x + y + z = 1$
Then:
:$x y + y z + z x < \dfrac 1 2$
=== Corollary ===
== Proof ==
We have: 
So:
:$2 \paren {x y + y z + z x} = 1 - \paren {x^2 + y^2 + z^2}$
We have: 
:$x^2 + y^2 + z^2 \ge 0$
for all real numbers $x$, $y$, $z$ with equality only if:
:$x = y = z = 0$
This cannot be the case since $x + y + z = 1$, so we have: 
:$x^2 + y^2 + z^2 > 0$
so:
:$2 \paren {x y + y z + z x} < 1$
giving:
:$x y + y z + z x < \dfrac 1 2$
== Sources ==
* : $1$. Say it with Letters: Algebraic Amusements: An Inequality
Category:Inequalities
Category:X + y + z equals 1 implies xy + yz + zx less than Half
X + y + z equals 1 implies xy + yz + zx less than Half/Corollary

== Theorem ==
Let $d$ be a non-zero real number. 
Let $x$, $y$ and $z$ be real numbers with: 
:$x + y + z = d$
Then: 
:$x y + y z + z x < \dfrac {d^2} 2$ 
== Proof ==
We have: 
:$\ds \frac x d + \frac y d + \frac z d = 1$
so, by $x + y + z = 1$ implies $x y + y z + z x < 1/2$, we have: 
:$\ds \frac x d \times \frac y d + \frac y d \times \frac z d + \frac z d \times \frac x d < 1$
That is: 
:$\ds \frac 1 {d^2} \paren {x y + y z + z x} < \frac 1 2$
giving:
:$\ds x y + y z + z x < \frac {d^2} 2$
Category:X + y + z equals 1 implies xy + yz + zx less than Half
X Choose n leq y Choose n + z Choose n-1 where n leq y leq x leq y+1 and n-1 leq z leq y

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $x, y \in \R$ be real numbers which satisfy:
:$n \le y \le x \le y + 1$
Let $z$ be the unique real number $z$ such that:
:$\dbinom x {n + 1} = \dbinom y {n + 1} + \dbinom z n$
where $n - 1 \le z \le y$.
Its uniqueness is proved at Uniqueness of Real $z$ such that $\dbinom x {n + 1} = \dbinom y {n + 1} + \dbinom z n$.
Then:
:$\dbinom x n \le \dbinom y n + \dbinom z {n - 1}$
== Proof ==
If $z \ge n$, then from Ordering of Binomial Coefficients:
:$\dbinom z {n + 1} \le \dbinom y {n + 1}$
Otherwise $n - 1 \le z \le n$, and:
:$\dbinom z {n + 1} \le 0 \le \dbinom y {n + 1}$
In either case:
:$(1): \quad \dbinom z {n + 1} \le \dbinom y {n + 1}$
Therefore:
and so $x \ge z + 1$.
Now we are to show that every term of the summation:
:$\ds \binom x {n + 1} - \binom y {n + 1} = \sum_{k \mathop \ge 0} \dbinom {z - k} {n - k} t_k$
where:
:$t_k = \dbinom {x - z - 1 + k} {k + 1} - \dbinom {y - z - 1 + k} {k + 1}$
is negative.
Because $z \ge n - 1$, the binomial coefficient $\dbinom {z - k} {n - k}$ is non-negative.
Because $x \ge z + 1$, the binomial coefficient $\dbinom {x - z - 1 + k} {k + 1}$ is also non-negative.
Therefore:
:$z \le y \le x$
implies that:
:$\dbinom {y - z - 1 + k} {k + 1} \le \dbinom {x - z - 1 + k} {k + 1}$
When $x = y$ and $z = n - 1$ the result becomes:
:$\dbinom x n \le \dbinom x n + \dbinom {n - 1} {n - 1}$
which reduces to:
:$\dbinom x n \le \dbinom x n + 1$
which is true.
Otherwise:
This is less than or equal to:
Hence the result.
== Sources ==
* : Problem $13.31 \ \text {(a)}$
* : $\S 1.2.6$: Binomial Coefficients: Exercise $66$
Category:Binomial Coefficients
X to the x is not of Exponential Order

== Theorem ==
Let $f: \R_{>0} \to \R$ be definedas:
:$\forall x \in \R_{>0}: \map f x = x^x$.
Then:
:$f$ is not of exponential order.
That is, it grows faster than any exponential.
== Proof ==
=== Lemma ===
By the definition of power:
:$\map f t = \map \exp {t \ln t}$ 
The theorem is equivalent to that there do not exist strictly positive real constants $M$, $K$, $a$ such that:
:$\forall t \ge M: \size {\map f t} < K e^{a t}$
 such constants $M$, $K$ and $a$ exist.
From the lemma, there exists a constant $C$ such that:
:$\forall t > C: \size {\map f t} > K e^{a t}$
For any $M$ chosen, $M + C > C$, thus from the lemma:
:$\size {\map f {M + C} } > K e^{a t}$
However, $M + C > M$, thus from the assumption:
:$\size {\map f {M + C} } < K e^{a t}$
Thus a contradiction has arisen.
It follows by Proof by Contradiction that our assumption was false.
Hence the result.
Category:Exponential Order
X to the x is not of Exponential Order/Lemma

== Lemma ==
Let $f: \R_{>0} \to \R$ be defined as:
:$\forall x \in \R_{>0}: \map f x = x^x$
Let there exist strictly positive real constants $M, K, a \in \R_{> 0}$ such that:
:$\forall t \ge M: \size {\map f t} < K e^{a t}$
Then there exists a constant $C$ such that:
:$\forall t > C: \size {\map f t} > K e^{a t}$
== Proof ==
By the definition of power:
:$\map f t = \map \exp {t \ln t}$ 
By Exponential of Real Number is Strictly Positive, we can reduce the lemma into the existence of $C$ such that:
:$\forall t > C: \map f t > K e^{a t}$
We will divide into two cases.
=== Case 1: $K > 1$ ===
Assume that $t > K e^a$.
Here, $C = K e^a$.
=== Case 2: $K \le 1$ ===
Assume that $t > e^a$.
Here:
:$C = e^a$
Category:Exponential Order
Xugu Zhaiqi Suanfa/Examples/Example 1

== Example of Problem from by ==
:Arrange the numbers $1$ to $33$ in these circles so that every circle and every diameter has the same total.
:
== Solution ==
:
The numbers in the circles add up to $138$.
The numbers along the diameters add up to $147$.
== Sources ==
* 
* 
* : Sun Tsu Suan-Ching: $75$
Category:Xugu Zhaiqi Suanfa
Y'-independent Euler's Equation

==Theorem==
Let $y$ be a mapping.
Let $J$ a functional be such that
:$\ds J \sqbrk y = \int_a^b \map F {x,y} \rd x$
Then the corresponding Euler's Equation can be reduced to:
:$F_y = 0$ 
Furthermore, this is an algebraic equation.
==Proof==
Assume that:
:$\ds J \sqbrk y = \int_a^b \map F {x,y} \rd x$
Then Euler's Equation for $J$ is:
:$F_y = 0$
Since $F$ is independent of $y'$, the equation is algebraic.
== Sources ==
* : $\S 1.4$: The Simplest Variational Problem. Euler's Equation
Category:Calculus of Variations
Yang-Mills Existence and Mass Gap

== Unsolved Problem ==
To establish rigorously:
:the Yang-Mills quantum theory
:the mass of the least massive particle of the force field is strictly positive
::(that is, the mass of each type of elementary particle is bounded below by a strictly positive value).
== Progress ==
== Sources ==
* 
* 
* : Appendix $18$: Millennium Prize problems: $4$.
* : Appendix $23$: Millennium Prize problems: $4$.
Category:Millennium Problems
Category:Unsolved Problems
Yff's Conjecture

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $\omega$ be the Brocard angle of $\triangle ABC$.
Then:
:$8 \omega^3 < ABC$
where $A, B, C$ are measured in radians.
== Proof ==
The Abi-Khuzam Inequality states that
:$\sin A \cdot \sin B \cdot \sin C \le \paren {\dfrac {3 \sqrt 3} {2 \pi} }^3 A \cdot B \cdot C$
The maximum value of $A B C - 8 \omega^3$ occurs when two of the angles are equal.
So taking $A = B$, and using $A + B + C = \pi$, the maximum occurs at the maximum of:
:$\map f A = A^2 \paren {\pi - 2 A} - 8 \paren {\map \arccot {2 \cot A - \cot 2 A} }^3$ 	
which occurs when:
:$2 A \paren {\pi - 3 A} - \dfrac {48 \paren {\map \arccot {\frac 1 2 \paren {3 \cot A + \tan A} } }^2 \paren {1 + 2 \cos 2 A} } {5 + 4 \cos 2 A} = 0$
== Also known as ==
Can also be seen referred to as the Yff conjecture.
== Historical Note ==
== Sources ==
* 
* 
* : $0,56559 56245 \ldots$
* 
* 
Category:Triangles
Yoneda Embedding Theorem

== Theorem ==
Let $C$ be a locally small category.
Let $\mathbf {Set}$ be the category of sets.
Let $\sqbrk {C^{\operatorname {op} }, \mathbf {Set} }$ be the contravariant functor category.
Then the Yoneda embedding $h_- : C \to \sqbrk {C^{\operatorname {op} }, \mathbf {Set} }$ is a fully faithful embedding.
== Proof ==
Category:Category Theory
Youden Square/Examples/Arbitrary Example 1

== Example of Youden Square ==
The following is an arbitrary example of a Youden square:
:$\begin{array} \mathrm A & \mathrm B & \mathrm C & \mathrm D & \mathrm E & \mathrm F & \mathrm G \\
\mathrm B & \mathrm E & \mathrm F & \mathrm G & \mathrm D & \mathrm A & \mathrm C \\
\mathrm C & \mathrm F & \mathrm D & \mathrm B & \mathrm A & \mathrm G & \mathrm E \\
\end {array}$
== Sources ==
* 
* 
Category:Examples of Youden Squares
Young's Inequality for Convolutions

== Theorem ==
Let $p, q, r \in \R_{\ge 1}$ satisfy:
:$1 + \dfrac 1 r = \dfrac 1 p + \dfrac 1 q$
Let $\map {L^p} {\R^n}$, $\map {L^q} {\R^n}$, and $\map {L^r} {\R^n}$ be Lebesgue spaces with seminorms $\norm {\, \cdot \,}_p$, $\norm {\, \cdot \,}_q$, and $\norm {\, \cdot \,}_r$ respectively.
Let $f \in \map {L^p} {\R^n}$ and $g \in \map {L^q} {\R^n}$.
Then the convolution $f * g$ is in $\map {L^r} {\R^n}$ and the following inequality is satisfied:
:$\norm {f * g}_r \le \norm f_p \cdot \norm g_q$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
We begin by seeking to bound $\size {\map {\paren {f * g} } x}$:
 
where the last inequality is via the Generalized Hölder's Inequality for Integrals applied to three functions.
Note that the relation of conjugate exponents in the Generalized Hölder's Inequality for Integrals is satisfied:
 
We now analyze terms $(1)$, $(2)$ and $(3)$ in turn:
Therefore we have
With these preliminary calculations out of the way, we turn to the main proof:
Category:Young's Inequality for Convolutions
Category:Measure Theory
Category:Inequalities
Young's Inequality for Convolutions/Corollary 1

== Corollary to Young's Inequality for Convolutions ==
Let $f: \R^n \to \R$ be a Lebesgue integrable function.
Let $p \in \R_{\ge 1}$.
Let $g: \R^n \to R$ be a Lebesgue $p$-integrable function.
Let $\norm f_p$ denote the $p$-seminorm of $f$.
Then the convolution $f * g$ of $f$ and $g$ satisfies:
:$\norm {f * g}_p \le \norm f_1 \cdot \norm g_p$
and hence is also Lebesgue $p$-integrable.
== Proof ==
Use Young's Inequality for Convolutions by letting $p'$, $q'$, and $r'$ be defined by $q' = r' = p$ and $p' = 1$.
== Sources ==
* : $14.6$
Category:Young's Inequality for Convolutions
Young's Inequality for Convolutions/Corollary 2

== Corollary to Young's Inequality for Convolutions ==
Let $f, g: \R^n \to \R$ be Lebesgue integrable functions.
Then their convolution $f * g$ is also Lebesgue integrable, and:
:$\norm {f * g} \le \norm f \norm g$
Thus, convolution may be seen as a binary operation:
:$*: \LL^1 \times \LL^1 \to \LL^1$
on the space of integrable functions $\LL^1$.
== Proof ==
Apply Young's Inequality for Convolutions with $p = q = r = 1$.
Category:Young's Inequality for Convolutions
Young's Inequality for Increasing Functions

== Theorem ==
Let $a_0$ and $b_0$ be strictly positive real numbers.
Let $f: \closedint 0 {a_0} \to \closedint 0 {b_0}$ be a strictly increasing bijection.
Let $a$ and $b$ be real numbers such that $0 \le a \le a_0$ and $0 \le b \le b_0$.
Then:
:$\ds ab \le \int_0^a \map f u \rd u + \int_0^b \map {f^{-1} } v \rd v$
where $\ds \int$ denotes the definite integral.
=== Equality ===
Equality holds in the above $b = \map f a$.
== Proof ==
== Also see ==
* Young's Inequality for Products
* Young's Inequality for Convolutions
Category:Young's Inequality for Increasing Functions
Category:Integral Calculus
Category:Strictly Increasing Real Functions
Young's Inequality for Increasing Functions/Equality

== Theorem ==
Let $a_0$ and $b_0$ be strictly positive real numbers.
Let $f: \closedint 0 {a_0} \to \closedint 0 {b_0}$ be a strictly increasing bijection.
Let $a$ and $b$ be real numbers such that $0 \le a \le a_0$ and $0 \le b \le b_0$.
Then $b = \map f a$ :
:$\ds a b = \int_0^a \map f u \rd u + \int_0^b \map {f^{-1} } v \rd v$
where $\ds \int$ denotes the Darboux integral.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
By Monotone Real Function is Darboux Integrable, $f$ and $f^{-1}$ are Darboux integrable.
Let $b = \map f a$.
Define:
:$\ds A = \int_0^a \map f u \rd u + \int_0^b \map {f^{-1} } v \rd v$
Consider any subdivision $P = \set {x_0, x_1, \ldots, x_n}$ of the closed real interval $\closedint 0 a$.
Then:
:$f P = \set {\map f {x_0}, \map f {x_1}, \ldots, \map f {x_n} }$
is a subdivision of $\closedint 0 b$.
We have that:
Similarly:
:$\map L {P, f} + \map U {f P, f^{-1} } = a b$
Let $\epsilon \in \R_{>0}$ be an arbitrary strictly positive real number.
By the definition of the Darboux integral, there exist subdivisions $P_+$ and $P_-$ of $\closedint 0 a$ such that:
:$a b = \map U {P_+, f} + \map L {f P_+, f^{-1} } < A + \epsilon$
:$a b = \map L {P_-, f} + \map U {f P_-, f^{-1} } > A - \epsilon$
The result follows from Real Plus Epsilon.
== Also see ==
* Young's Inequality for Products
* Young's Inequality for Convolutions
Category:Young's Inequality for Increasing Functions
Young's Inequality for Products

== 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 :
:$b = a^{p - 1}$
== Proof by Convexity ==
== Geometric Proof ==
== Proof by Calculus ==
== Parameter Inequalities ==
== Sources ==
* : Exercise $6.10 a$
* 
* 
Category:Young's Inequality for Products
Category:Real Analysis
Category:Inequalities
Young's Inequality for Products/Geometric Proof

== Theorem ==
== Proof ==
:
In the above diagram:
:the region corresponds to $\ds \int_0^\alpha t^{p - 1} \rd t$
:the 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 $q > 1$.
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.
It remains to prove the equality condition:
:$a b = \dfrac {a^p} p + \dfrac {b^q} q$
:
:$b = a^{p - 1}$
=== Necessary Condition ===
Let $b = a^{p - 1}$.
Then:
That is:
:$b = a^{p - 1} \implies a b = \dfrac {a^p} p + \dfrac {b^q} q$
=== Sufficient Condition ===
Let:
:$a b = \dfrac {a^p} p + \dfrac {b^q} q$
 $b \ne a^{p - 1}$.
==== 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 and 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 $u$ 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:
:$a b = \dfrac {a^p} p + \dfrac {b^q} q \implies b = a^{p - 1}$
== Sources ==
* : $12.1$
Category:Young's Inequality for Products
Young's Inequality for Products/Parameter Inequalities

== Theorem ==
Statements of Young's Inequality for Products will commonly insist that $p, q > 1$.
However, from Positive Real Numbers whose Reciprocals Sum to 1 we have that if:
:$p, q > 0$
and:
:$\dfrac 1 p + \dfrac 1 q = 1$ 
it follows directly that $p, q > 1$.
== Sources ==
* 
* 
Category:Young's Inequality for Products
Young's Inequality for Products/Proof by Calculus

== Theorem ==
== Proof ==
 assume that $a^p \ge b^q$, otherwise swap $a$ with $b$ and $p$ with $q$. 
Define $f : \hointr 0 \infty \to \R$ by: 
:$\ds \map f t = \frac {t^p} p + \frac 1 q - t$
We have, from Derivative of Power and Sum Rule for Derivatives:
:$\ds \map {f'} t = t^{p - 1} - 1$
So for $t \ge 1$ we have: 
:$\ds \map {f'} t \ge 0$
So, from Real Function with Positive Derivative is Increasing, we have: 
:$f$ is increasing on $\hointr 1 \infty$. 
That is: 
:$\map f t \ge \map f 1$
for all $t \ge 1$.
Since $a^p \ge b^q$, we have: 
:$a^p b^{-q} \ge 1$
so:
:$a b^{-q/p} \ge 1$
So:
:$\map f {a b^{-q/p} } \ge \map f 1$
That is: 
:$\ds \frac {a^p b^{-q} } p + \frac 1 q - a b^{-q/p} \ge \frac 1 p + \frac 1 q - 1$
By hypothesis, we have: 
:$\ds \frac 1 p + \frac 1 q = 1$
So it follows that: 
:$\ds \frac {a^p b^{-q} } p + \frac 1 q \ge a b^{-q/p}$
So:
:$\ds \frac {a^p} p + \frac {b^q} q \ge a b^{q \paren {1 - 1/p} }$
Finally, we have: 
:$\ds 1 - \frac 1 p = \frac 1 q$
so:
:$\ds q \paren {1 - \frac 1 p} = 1$
giving: 
:$\ds \frac {a^p} p + \frac {b^q} q \ge a b$
For the equality case, note that: 
:$\map {f'} t > 0$
for $t > 1$.
So, from Real Function with Strictly Positive Derivative is Strictly Increasing, we have: 
:$f$ is strictly increasing for $t > 1$.
So: 
:$\map f {a b^{-q/p} } = \map f 1$
that is: 
:$\ds a b = \frac {a^p} p + \frac {b^q} q$
: 
:$a b^{-q/p} = 1$ 
That is:
:$b = a^{p/q}$
We have:
:$\ds \frac 1 p + \frac 1 q = 1$
and so:
:$\ds 1 + \frac p q = p$
giving:
:$\ds \frac p q = p - 1$
So we have:
:$b = a^{p - 1}$
as required.
Category:Young's Inequality for Products