Difference between revisions of "TPA's equality"

(Go ask tpa for clarification before labelling it as "fake)
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'''TPA's equality''', also known as the TPA-snek equality, is an equality discovered by AoPS user thepiercingarrow in 2017, said to have been inspired from doing calculus with snek.
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− '''DISCLAIMER: THIS IS DEFINITELY A REAL EQUALITY AND YOU CAN TOTALLY USE THIS ON THE AMO NEXT YEAR'''
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− ==Inequality==
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− Let <math>F</math> be any symmetric inequality involving positive numbers. Let <math>x_1, \cdots,x_n</math> be the parts of that inequality. In contest space, if <math>F</math> achieves the equality then it is for <math>x_1 = \cdots = x_n.</math>
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− ==Proof==
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− The proof of TPA's equality is very complicated. It requires using quantum computing to search all inequalities over contest space and checking the verity of the TPA equality for each one. A program was developed by alifenix- in TPA-Snek Labs in Equestria, and ran on the TPA-Snek quantum computer. This program is now open-sourced and available for anyone to run. It can be found on the TPA-Snek Labs main website.
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− ==Problems==
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− ===Introductory===
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− Prove the equality case of AM-GM using TPA's equality.
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− ===Intermediate===
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− * Prove that for any <math>\triangle ABC</math>, we have the maximum of <math>\sin{A}+\sin{B}+\sin{C}</math> is <math>\frac{3\sqrt{3}}{2}</math>.
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− ===Olympiad===
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− *Let <math>a,b,c</math> be positive real numbers. Prove that the minimum of <math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}</math> is <math>1</math> ([[2001 IMO Problems/Problem 2|Source]])
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− ==Extension to all reals==
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− *Consider <math>x^4+y^4 - 2 (x^2 +y^2) + x y \ge-\frac{25}{8}</math>. When x is set equal to y then there is no solution for x and y that satisfies the equality. However when taking <math>(x,y)=\left(\frac{\sqrt{5}}{2},-\frac{\sqrt{5}}{2} \right)</math> the equality does hold. This leads us to think that given a 2 var inequality for all reals, we find the equality case when the variables are additive inverses.
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− ==Clarification of Domain of TPA's equality==
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− *Consider <math>x^4+y^4 - 2 (x^2 +y^2) + x y+4 x^2 y^2 \ge-\frac{9}{8}</math>. When x is set equal to y then there is no solution for x and y that satisfies the equality. However when taking <math>(x,y)=\left( \frac{1}{4}\left(\sqrt{2}-\sqrt{6}\right), \frac{1}{4}\left(\sqrt{2}+\sqrt{6}\right) \right)</math> the equality does hold so it is a strict equality. This is because <math>- 2 (x^2 +y^2)</math> is not positive, and thus this does not count as an inequality involving positive parts.

Revision as of 17:27, 25 October 2018

TPA's equality, also known as the TPA-snek equality, is an equality discovered by AoPS user thepiercingarrow in 2017, said to have been inspired from doing calculus with snek.


DISCLAIMER: THIS IS DEFINITELY A REAL EQUALITY AND YOU CAN TOTALLY USE THIS ON THE AMO NEXT YEAR

− ==Inequality==

− Let $F$ be any symmetric inequality involving positive numbers. Let $x_1, \cdots,x_n$ be the parts of that inequality. In contest space, if $F$ achieves the equality then it is for $x_1 = \cdots = x_n.$

− ==Proof==

− The proof of TPA's equality is very complicated. It requires using quantum computing to search all inequalities over contest space and checking the verity of the TPA equality for each one. A program was developed by alifenix- in TPA-Snek Labs in Equestria, and ran on the TPA-Snek quantum computer. This program is now open-sourced and available for anyone to run. It can be found on the TPA-Snek Labs main website.

− ==Problems==

− ===Introductory===

− Prove the equality case of AM-GM using TPA's equality.

− ===Intermediate===

− * Prove that for any $\triangle ABC$, we have the maximum of $\sin{A}+\sin{B}+\sin{C}$ is $\frac{3\sqrt{3}}{2}$.

− ===Olympiad===

− *Let $a,b,c$ be positive real numbers. Prove that the minimum of $\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}$ is $1$ (Source)

− ==Extension to all reals==

− *Consider $x^4+y^4 - 2 (x^2 +y^2) + x y \ge-\frac{25}{8}$. When x is set equal to y then there is no solution for x and y that satisfies the equality. However when taking $(x,y)=\left(\frac{\sqrt{5}}{2},-\frac{\sqrt{5}}{2} \right)$ the equality does hold. This leads us to think that given a 2 var inequality for all reals, we find the equality case when the variables are additive inverses.

− ==Clarification of Domain of TPA's equality==

− *Consider $x^4+y^4 - 2 (x^2 +y^2) + x y+4 x^2 y^2 \ge-\frac{9}{8}$. When x is set equal to y then there is no solution for x and y that satisfies the equality. However when taking $(x,y)=\left( \frac{1}{4}\left(\sqrt{2}-\sqrt{6}\right), \frac{1}{4}\left(\sqrt{2}+\sqrt{6}\right) \right)$ the equality does hold so it is a strict equality. This is because $- 2 (x^2 +y^2)$ is not positive, and thus this does not count as an inequality involving positive parts.

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