In AoPS' (Art of Problem Solving) proof of Muirhead's inequality, how does the below equality work o

Theresa Archer

Theresa Archer

Answered question

2022-06-06

In AoPS' (Art of Problem Solving) proof of Muirhead's inequality, how does the below equality work out?
The below equation appears to show two expressions (1 and 2), each under the symmetric sum notation, being multiplied together to produce expression 3, which is also under the symmetric sum notation...
but normally, [ A 1 + A 2 ( ) + A n ] [ B 1 + B 2 + ( ) + B n ] is not [ A 1 B 1 + A 2 B 2 + ( ) + A n B n ] ,
that is, when we multiply the sum of A i ( i = [ 1 ,   n ] ) and B i ( i = [ 1 ,   n ] ), the result is not the sum of A i B i ( i = [ 1 ,   n ] )

Answer & Explanation

Punktatsp

Punktatsp

Beginner2022-06-07Added 22 answers

Step 1
Yes, that proof is bogus. If you examine it, they only use the fact that a i = b i and not the other conditions of majorization. So this "proof" purports to show that if a i = b i then
σ S n x σ ( 1 ) a 1 x σ ( n ) a n σ S n x σ ( 1 ) b 1 x σ ( n ) b n . .
(And hence they are equal, as well.) I mean this just can't be true for trivial reasons.

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