TPA's equality

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 FAKE

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 $\frac{3\sqrt{3}}{2}$.

• 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)

Extention 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.

Counterexample

The minimum of the polynomial $((x-2)^2+(y-1)^2)*((x-1)^2+(y-2)^2))$ cannot exceed 0 by trivial inequality. It is a symmetrical polynomial where each part is positive by trivial inequality. It is also achieved at $(x,y)=(1, 2)$ or $(x,y)=(2, 1)$, where the polynomial has a value of 0. For more information about this topic, read: https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Waterhouse378-387.pdf