Well by looking at it it seems pretty trivial...

Trivial by Muirhead and AM-GM.

asdf334 wrote: Trivial by Muirhead and AM-GM. This proof is incorrect. This is a very difficult 25 MOHS inequality!

DottedCaculator wrote: Prove that $a^3 + b^3 + c^3 + abc +a^{3}b^{2}c^{-1}+a^{3}c^{2}b^{-1}+b^{3}a^{2}c^{-1}+b^{3}c^{2}a^{-1}+c^{3}a^{2}b^{-1}+c^{3}b^{2}a^{-1}+a^{5}b^{3}c^{-3}+ abc^{14} + a^{5}c^{3}b^{-3}+b^{5}a^{3}c^{-3}+b^{5}c^{3}a^{-3}+c^{5}a^{3}b^{-3}+c^{5}b^{3}a^{-3}+a^{6}b^{1}c^{-1}+a^{6}c^{1}b^{-1}+b^{6}a^{1}c^{-1}+b^{6}c^{1}a^{-1}+c^{6}a^{1}b^{-1}+c^{6}b^{1}a^{-1}+ a^{6}b^{4}c^{-3}+a^{6}c^{4}b^{-3}+b^{6}a^{4}c^{-3}+b^{6}c^{4}a^{-3}+c^{6}a^{4}b^{-3}+c^{6}b^{4}a^{-3}+a^{7}b^{2}c^{-1}+a^{7}c^{2}b^{-1}+b^{7}a^{2}c^{-1}+b^{7}c^{2}a^{-1}+c^{7}a^{2}b^{-1}+ abc + a^{14}bc + c^{7}b^{2}a^{-1}+a^{4}b^{1}c^{4}+a^{4}c^{1}b^{4}+b^{4}a^{1}c^{4}+b^{4}c^{1}a^{4}+c^{4}a^{1}b^{4}+c^{4}b^{1}a^{4}+a^{6}c^{4}+a^{6}b^{4}+b^{6}c^{4}+b^{6}a^{4}+c^{6}b^{4}+c^{6}a^{4}+a^{9}b^{6}c^{-4}+a^{9}c^{6}b^{-4}+ ab^{14}c + b^{9}a^{6}c^{-4}+b^{9}c^{6}a^{-4}+c^{9}a^{6}b^{-4}+ abc + c^{9}b^{6}a^{-4}+a^{12}b^{1}c^{-1}+a^{12}c^{1}b^{-1}+b^{12}a^{1}c^{-1}+b^{12}c^{1}a^{-1}+c^{12}a^{1}b^{-1}+ c^5 b^5 a^5 - c^5 b^5 a^2 + 3 c^5 b^5 - c^5 b^2 a^5 + c^5 b^2 a^2 - 3 c^5 b^2 + 3 c^5 a^5 - 3 c^5 a^2 + 9 c^5 - c^2 b^5 a^5 + c^2 b^5 a^2 - 3 c^2 b^5 + c^2 b^2 a^5 - c^2 b^2 a^2 + 3 c^2 b^2 - 3 c^2 a^5 + 3 c^2 a^2 - 9 c^2 + 3 b^5 a^5 - 3 b^5 a^2 + 9 b^5 - 3 b^2 a^5 + 3 b^2 a^2 - 9 b^2 + 9 a^5 - 9 a^2 + 27 + c^{12}b^{1}a^{-1}+a^{13}b^{9}c^{-9}+a^{13}c^{9}b^{-9}+b^{13}a^{9}c^{-9}+b^{13}c^{9}a^{-9}+c^{13}a^{9}b^{-9}+c^{13}b^{9}a^{-9}+a^{12}b^{11}c^{-9}+a^{12}c^{11}b^{-9}+b^{12}a^{11}c^{-9}+b^{12}c^{11}a^{-9}+c^{12}a^{11}b^{-9}+c^{12}b^{11}a^{-9}+a^{8}b^{7}+a^{8}c^{7}+b^{8}a^{7}+b^{8}c^{7}+c^{8}a^{7}+c^{8}b^{7} + a^{16} + b^{16} + c^{16} + a^{16} + b^{16} + c^{16} + a^{16} + b^{16} + c^{16}\ge c^3 + 3 c^2 a + 3 c b^2 + 6 c b a + b^3 + 3 b^2 a + a^3 + a^{1}c^{2}+a^{1}b^{2}+4b^{1}c^{2}+4b^{1}a^{2}+c^{1}b^{2}+4c^{1}a^{2}+a^{1}c^{3}+a^{1}b^{3}+b^{1}c^{3}+b^{1}a^{3}+c^{1}b^{3}+c^{1}a^{3}+a^{3}b^{2}+a^{3}c^{2}+b^{3}a^{2}+b^{3}c^{2}+c^{3}a^{2}+c^{3}b^{2}+a^{5}c^{1}+a^{5}b^{1}+b^{5}c^{1}+b^{5}a^{1}+c^{5}b^{1}+c^{5}a^{1}+a^{2}b^{1}c^{4}+a^{2}c^{1}b^{4}+b^{2}a^{1}c^{4}+b^{2}c^{1}a^{4}+c^{2}a^{1}b^{4}+c^{2}b^{1}a^{4}+a^{1}c^{7}+a^{1}b^{7}+b^{1}c^{7}+b^{1}a^{7}+c^{1}b^{7}+c^{1}a^{7}+a^{1}c^{8}+a^{1}b^{8}+b^{1}c^{8}+b^{1}a^{8}+c^{1}b^{8}+c^{1}a^{8}+a^{5}b^{1}c^{4}+a^{5}c^{1}b^{4}+b^{5}a^{1}c^{4}+b^{5}c^{1}a^{4}+c^{5}a^{1}b^{4}+c^{5}b^{1}a^{4}+a^{2}b^{1}c^{8}+a^{2}c^{1}b^{8}+b^{2}a^{1}c^{8}+b^{2}c^{1}a^{8}+c^{2}a^{1}b^{8}+c^{2}b^{1}a^{8}+a^{1}c^{11}+a^{1}b^{11}+b^{1}c^{11}+b^{1}a^{11}+c^{1}b^{11}+c^{1}a^{11}+a^{6}b^{2}c^{5}+a^{6}c^{2}b^{5}+b^{6}a^{2}c^{5}+b^{6}c^{2}a^{5}+c^{6}a^{2}b^{5}+c^{6}b^{2}a^{5}+a^{3}b^{2}c^{9}+a^{3}c^{2}b^{9}+b^{3}a^{2}c^{9}+b^{3}c^{2}a^{9}+c^{3}a^{2}b^{9}+c^{3}b^{2}a^{9}+a^{3}b^{1}c^{11}+a^{3}c^{1}b^{11}+b^{3}a^{1}c^{11}+b^{3}c^{1}a^{11}+c^{3}a^{1}b^{11}+c^{3}b^{1}a^{11} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15}c^{2}a^{1}b^{4}+c^{2}b^{1}a^{4}+a^{1}c^{7}+a^{1}b^{7}+b^{1}c^{7}+b^{1}a^{7}+c^{1}b^{7}+c^{1}a^{7}+a^{1}c^{8}+a^{1}b^{8}+b^{1}c^{8}+b^{1}a^{8}+c^{1}b^{8}+c^{1}a^{8}+a^{5}b^{1}c^{4}+a^{5}c^{1}b^{4}+b^{5}a^{1}c^{4}+b^{5}c^{1}a^{4}+c^{5}a^{1}b^{4}+c^{5}b^{1}a^{4}+a^{2}b^{1}c^{8}+a^{2}c^{1}b^{8}+b^{2}a^{1}c^{8}+b^{2}c^{1}a^{8}+c^{2}a^{1}b^{8}+c^{2}b^{1}a^{8}+a^{1}c^{11}+a^{1}b^{11}+b^{1}c^{11}+b^{1}a^{11}+c^{1}b^{11}+c^{1}a^{11}+a^{6}b^{2}c^{5}+a^{6}c^{2}b^{5}+b^{6}a^{2}c^{5}+b^{6}c^{2}a^{5}+c^{6}a^{2}b^{5}+c^{6}b^{2}a^{5}+a^{3}b^{2}c^{9}+a^{3}c^{2}b^{9}+b^{3}a^{2}c^{9}+b^{3}c^{2}a^{9}+c^{3}a^{2}b^{9}+c^{3}b^{2}a^{9}+a^{3}b^{1}c^{11}+a^{3}c^{1}b^{11}+b^{3}a^{1}c^{11}+b^{3}c^{1}a^{11}+c^{3}a^{1}b^{11}+c^{3}b^{1}a^{11} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15}$ for all $a,b,c\in\mathbb R^+$. Proposed by Henry Jiang and C++ I don't even know what that says, DottedCaculator(Rating 100 so I expect you always do hard problems)

Trivial by 16th degree schur and holder.

This is also incorrect.

Prove me wrong.

Trivial by ISL 2019 A6 - after an application of the Hilbert nullstellensatz and the projective nullstellensatz, you get a simple expression which can be proven true by the Siefert van-kampen theorem as this implies that 1+1 = 2 which is true!

carvilesp7 wrote: Trivial by ISL 2019 A6 - after an application of the Hilbert nullstellensatz and the projective nullstellensatz, you get a simple expression which can be proven true by the Siefert van-kampen theorem as this implies that 1+1 = 2 which is true! I like your funny words, magic man

Unfortunately, this proof is also incorrect. Your polynomial does not form a bijection from this inequality to $1+1=2$.

Trivial by Wolfram-Alpha.

asdf334 wrote: Trivial by Wolfram-Alpha. Oops you've exceeded the maximum number of characters.

LHS >= RHS because if not, the statement is wrong.

In fact, the statement was wrong until Holden pointed out a mistake.

you conveniently grouped terms together so its easier for me to apply muirheads! thanks!

Incorrect. Notice that the left hand side has an $abc$ term which cannot be eliminated by Muirhead.

you probably took some isl problem and added some random muirhead to disguise it

This is also incorrect.

I don't even see \leq or \geq. Anyway where are all RELSMO problems?

i am certainly finding it very hard to memorize the problem statement

DottedCaculator wrote: This is a very difficult 25 MOHS inequality! meaning you took some 25M inequality and added random spam

carvilesp7 wrote: Trivial by ISL 2019 A6 - after an application of the Hilbert nullstellensatz and the projective nullstellensatz, you get a simple expression which can be proven true by the Siefert van-kampen theorem as this implies that 1+1 = 2 which is true! Simplifying the inequality presented, we get $2 \geq 1$. We use the above claim to finish the problem: Since $2=1+1$, we substitute + simplify to get $1 \geq 0$. Claim: $1 \geq 0$ Proof: I have 10 fingers. If I cut 9 off, I will be left with 1. If I cut 10 off, I will be left with none. Case 1: I will have 1 finger, and will still be able to use it. Case 2: I will not have any fingers, and will not be able to do anything with my remaining fingers (none). $\square$ Thus we can form a bijection between finger usability and $1 \geq 0$, which bridges the gap and shows the bijection between $1+1=2$ and the given inequality. $\blacksquare$

DottedCaculator wrote: The solution is clean and conceptual. The problem statement is memorable. There is an unexpected flavor. Every time I read the problem I can’t help nodding my head and laughing … The problem resembles wisdom, ingenuity, and best of all, a great sense of humor.

Taco12 wrote: carvilesp7 wrote: Trivial by ISL 2019 A6 - after an application of the Hilbert nullstellensatz and the projective nullstellensatz, you get a simple expression which can be proven true by the Siefert van-kampen theorem as this implies that 1+1 = 2 which is true! Simplifying the inequality presented, we get $2 \geq 1$. We use the above claim to finish the problem: Since $2=1+1$, we substitute + simplify to get $1 \geq 0$. Claim: $1 \geq 0$ Proof: I have 10 fingers. If I cut 9 off, I will be left with 1. If I cut 10 off, I will be left with none. Case 1: I will have 1 finger, and will still be able to use it. Case 2: I will not have any fingers, and will not be able to do anything with my remaining fingers (none). $\square$ Thus we can form a bijection between finger usability and $1 \geq 0$, which bridges the gap and shows the bijection between $1+1=2$ and the given inequality. $\blacksquare$ This conjecture has been proposed by Albert de la Fetti in 1653! This is a historic moment, folks. I am glad to be one of the few people to witness his genius during his lifetime.

just assume a,b,c=1 and count up the terms

Can enyone answer the question

Certainly one of the problems of all time.

ok time for the solution

DottedCaculator wrote: ok time for the solution

$$(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\geq(a+b+c)^3$$USAMO 2004

TYT wrote: Certainly one of the problems of all time. Bravo, of course

Also check out the other ELSMO revenge problems!

DottedCaculator wrote: Also check out the other ELSMO revenge problems! ok

why is there a random $c^{2}a^{1}b^{4}+c^{2}b^{1}a^{4}$ there

DottedCaculator wrote: Prove that $a^3 + b^3 + c^3 + abc +a^{3}b^{2}c^{-1}+a^{3}c^{2}b^{-1}+b^{3}a^{2}c^{-1}+b^{3}c^{2}a^{-1}+c^{3}a^{2}b^{-1}+c^{3}b^{2}a^{-1}+a^{5}b^{3}c^{-3}+ abc^{14} + a^{5}c^{3}b^{-3}+b^{5}a^{3}c^{-3}+b^{5}c^{3}a^{-3}+c^{5}a^{3}b^{-3}+c^{5}b^{3}a^{-3}+a^{6}b^{1}c^{-1}+a^{6}c^{1}b^{-1}+b^{6}a^{1}c^{-1}+b^{6}c^{1}a^{-1}+c^{6}a^{1}b^{-1}+c^{6}b^{1}a^{-1}+ a^{6}b^{4}c^{-3}+a^{6}c^{4}b^{-3}+b^{6}a^{4}c^{-3}+b^{6}c^{4}a^{-3}+c^{6}a^{4}b^{-3}+c^{6}b^{4}a^{-3}+a^{7}b^{2}c^{-1}+a^{7}c^{2}b^{-1}+b^{7}a^{2}c^{-1}+b^{7}c^{2}a^{-1}+c^{7}a^{2}b^{-1}+ abc + a^{14}bc + c^{7}b^{2}a^{-1}+a^{4}b^{1}c^{4}+a^{4}c^{1}b^{4}+b^{4}a^{1}c^{4}+b^{4}c^{1}a^{4}+c^{4}a^{1}b^{4}+c^{4}b^{1}a^{4}+a^{6}c^{4}+a^{6}b^{4}+b^{6}c^{4}+b^{6}a^{4}+c^{6}b^{4}+c^{6}a^{4}+a^{9}b^{6}c^{-4}+a^{9}c^{6}b^{-4}+ ab^{14}c + b^{9}a^{6}c^{-4}+b^{9}c^{6}a^{-4}+c^{9}a^{6}b^{-4}+ abc + c^{9}b^{6}a^{-4}+a^{12}b^{1}c^{-1}+a^{12}c^{1}b^{-1}+b^{12}a^{1}c^{-1}+b^{12}c^{1}a^{-1}+c^{12}a^{1}b^{-1}+ c^5 b^5 a^5 - c^5 b^5 a^2 + 3 c^5 b^5 - c^5 b^2 a^5 + c^5 b^2 a^2 - 3 c^5 b^2 + 3 c^5 a^5 - 3 c^5 a^2 + 9 c^5 - c^2 b^5 a^5 + c^2 b^5 a^2 - 3 c^2 b^5 + c^2 b^2 a^5 - c^2 b^2 a^2 + 3 c^2 b^2 - 3 c^2 a^5 + 3 c^2 a^2 - 9 c^2 + 3 b^5 a^5 - 3 b^5 a^2 + 9 b^5 - 3 b^2 a^5 + 3 b^2 a^2 - 9 b^2 + 9 a^5 - 9 a^2 + 27 + c^{12}b^{1}a^{-1}+a^{13}b^{9}c^{-9}+a^{13}c^{9}b^{-9}+b^{13}a^{9}c^{-9}+b^{13}c^{9}a^{-9}+c^{13}a^{9}b^{-9}+c^{13}b^{9}a^{-9}+a^{12}b^{11}c^{-9}+a^{12}c^{11}b^{-9}+b^{12}a^{11}c^{-9}+b^{12}c^{11}a^{-9}+c^{12}a^{11}b^{-9}+c^{12}b^{11}a^{-9}+a^{8}b^{7}+a^{8}c^{7}+b^{8}a^{7}+b^{8}c^{7}+c^{8}a^{7}+c^{8}b^{7} + a^{16} + b^{16} + c^{16} + a^{16} + b^{16} + c^{16} + a^{16} + b^{16} + c^{16}\ge c^3 + 3 c^2 a + 3 c b^2 + 6 c b a + b^3 + 3 b^2 a + a^3 + a^{1}c^{2}+a^{1}b^{2}+4b^{1}c^{2}+4b^{1}a^{2}+c^{1}b^{2}+4c^{1}a^{2}+a^{1}c^{3}+a^{1}b^{3}+b^{1}c^{3}+b^{1}a^{3}+c^{1}b^{3}+c^{1}a^{3}+a^{3}b^{2}+a^{3}c^{2}+b^{3}a^{2}+b^{3}c^{2}+c^{3}a^{2}+c^{3}b^{2}+a^{5}c^{1}+a^{5}b^{1}+b^{5}c^{1}+b^{5}a^{1}+c^{5}b^{1}+c^{5}a^{1}+a^{2}b^{1}c^{4}+a^{2}c^{1}b^{4}+b^{2}a^{1}c^{4}+b^{2}c^{1}a^{4}+c^{2}a^{1}b^{4}+c^{2}b^{1}a^{4}+a^{1}c^{7}+a^{1}b^{7}+b^{1}c^{7}+b^{1}a^{7}+c^{1}b^{7}+c^{1}a^{7}+a^{1}c^{8}+a^{1}b^{8}+b^{1}c^{8}+b^{1}a^{8}+c^{1}b^{8}+c^{1}a^{8}+a^{5}b^{1}c^{4}+a^{5}c^{1}b^{4}+b^{5}a^{1}c^{4}+b^{5}c^{1}a^{4}+c^{5}a^{1}b^{4}+c^{5}b^{1}a^{4}+a^{2}b^{1}c^{8}+a^{2}c^{1}b^{8}+b^{2}a^{1}c^{8}+b^{2}c^{1}a^{8}+c^{2}a^{1}b^{8}+c^{2}b^{1}a^{8}+a^{1}c^{11}+a^{1}b^{11}+b^{1}c^{11}+b^{1}a^{11}+c^{1}b^{11}+c^{1}a^{11}+a^{6}b^{2}c^{5}+a^{6}c^{2}b^{5}+b^{6}a^{2}c^{5}+b^{6}c^{2}a^{5}+c^{6}a^{2}b^{5}+c^{6}b^{2}a^{5}+a^{3}b^{2}c^{9}+a^{3}c^{2}b^{9}+b^{3}a^{2}c^{9}+b^{3}c^{2}a^{9}+c^{3}a^{2}b^{9}+c^{3}b^{2}a^{9}+a^{3}b^{1}c^{11}+a^{3}c^{1}b^{11}+b^{3}a^{1}c^{11}+b^{3}c^{1}a^{11}+c^{3}a^{1}b^{11}+c^{3}b^{1}a^{11} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15}+c^{2}a^{1}b^{4}+c^{2}b^{1}a^{4}+a^{1}c^{7}+a^{1}b^{7}+b^{1}c^{7}+b^{1}a^{7}+c^{1}b^{7}+c^{1}a^{7}+a^{1}c^{8}+a^{1}b^{8}+b^{1}c^{8}+b^{1}a^{8}+c^{1}b^{8}+c^{1}a^{8}+a^{5}b^{1}c^{4}+a^{5}c^{1}b^{4}+b^{5}a^{1}c^{4}+b^{5}c^{1}a^{4}+c^{5}a^{1}b^{4}+c^{5}b^{1}a^{4}+a^{2}b^{1}c^{8}+a^{2}c^{1}b^{8}+b^{2}a^{1}c^{8}+b^{2}c^{1}a^{8}+c^{2}a^{1}b^{8}+c^{2}b^{1}a^{8}+a^{1}c^{11}+a^{1}b^{11}+b^{1}c^{11}+b^{1}a^{11}+c^{1}b^{11}+c^{1}a^{11}+a^{6}b^{2}c^{5}+a^{6}c^{2}b^{5}+b^{6}a^{2}c^{5}+b^{6}c^{2}a^{5}+c^{6}a^{2}b^{5}+c^{6}b^{2}a^{5}+a^{3}b^{2}c^{9}+a^{3}c^{2}b^{9}+b^{3}a^{2}c^{9}+b^{3}c^{2}a^{9}+c^{3}a^{2}b^{9}+c^{3}b^{2}a^{9}+a^{3}b^{1}c^{11}+a^{3}c^{1}b^{11}+b^{3}a^{1}c^{11}+b^{3}c^{1}a^{11}+c^{3}a^{1}b^{11}+c^{3}b^{1}a^{11} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15}$ for all $a,b,c\in\mathbb R^+$. Proposed by Henry Jiang and C++ Splendid and correct, as customary