Reversing the Arithmetic mean Geometric mean inequality

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1 Reversig the Arithmetic me Geometric me iequlity Tie Lm Nguye Abstrct I this pper we discuss some iequlities which re obtied by ddig o-egtive expressio to oe of the sides of the AM-GM iequlity I this wy we c reverse the AM-GM iequlity Itroductio The AM-GM iequlity is oe of the best iequlities d it is very useful for provig iequlities We will recll the AM-GM iequlity s follows Theorem For ll o-egtive rel umbers we hve ) where is turl umber greter th Equlity holds if d oly if = = = Now we will observe questio: How c we reverse AM-GM iequlity? The first thig tht sprigs to mid is ddig o-egtive expressio to the right-hd-side of the AM-GM iequlity It is turl tht the obtied iequlity becomes equlity if d oly if AM-GM iequlity becomes equlity tht is if we dd expressio M : = f ) M is fuctio of vribles ) to the right-hd-side of AM-GM iequlity the M must stisfy the two followig coditios simulteously i) M : = f ) 0 for o-egtive umbers ii) M : = f ) = 0 if d oly if = = = So we c choose my expressios such s: M = k i ; M = k i i + M = k mx or < i M4 = k4 i ) etc < i with the coditio ) + ; { } i= < i The uthor thks Thh H Le for her help i the preprtio of this pper

2 Two iequlities Propositio Give turl umber The the smllest rel umber k such tht for ll o-egtive umbers we hve the followig iequlity k i ) is < i Proof Suppose tht ) is stisfied for ll 0 Try = = = > 0 d = 0 from ) we deduce k Now we will prove tht ) is correct whe k = ie < i i for ll o-egtive rel umbers Without losig geerlity we ssume tht Thus the iequlity we eed to prove is equivlet to ) + ) + 5) ) ) + + or ) + where [ x ] is the lrgest iteger which is smller th x It is esy to see tht the lst iequlity is correct sice The proof is completed Corollry For ll o-egtive rel umbers we lwys hve mx { i } ) < i! where =! )! Proof For i we hve terms i d ote tht i is ot greter th mx ) follows immeditely from ) < i { i }

3 + + + Corollry 4 + i + 4) i i= for ll rel umbers 0 Proof Usig the iequlities i ) i i+ i+ i+ for ll pirs i stisfyig i we c esily deduce tht i ) i i+ Thus from ) we deduce 4) < i i= Propositio 5 Give turl umber greter th the smllest rel umber k such tht for ll o-egtive umbers we hve the iequlity k mx { i } 5) < i is Proof Sice 5) is stisfied for ll rel umbers 0 tryig = = = > 0 d = 0 from 5) we deduce k We will prove tht 5) is correct whe k = tht is mx { i } for ll o-egtive rel umbers < i Without losig geerlity we ssume tht 0 The < i { i } mx = Hece the iequlity which eeds be proved is equivlet to ) ) or ) + We hve ) d Therefore sice This completes the proof Corollry 6 For ll o-egtive rel umbers we hve i i + 6) i= with the coditio + Proof Suppose tht mx { i } i = Usig the iequlity from 5) we get i i+ i= O the other hd we hve +

4 Addig two bove iequlities side by side we get i i+ i= or + i i + i= Corollry 7 For ll o-egtive rel umbers we hve i + ) i i= with the coditio + Proof By usig the Cuchy Schwrz iequlity i i+ i i+ i= i= b i i i bi i= i= i= we hve So from this d 6) we immeditely hve 7) 7) Aother proof of ) i the cse = I this sectio we will show ew proof of ) for the cse = Lemm 8 For ll o-egtive rel umbers + b + b 8 b we hve + b 8) Proof Without losig geerlity we ssume tht b 0 The 8) is equivlet to the followig iequlity + b + b + b ) or b + b) 6b This is 8 correct becuse b 0 The lemm is proved Problem 9 For ll o-egtive rel umbers we hve ) 9) Proof We itroduce the followig ottios: b = b = b = the b b b 0 b + b + b d 9) becomes bb b+ b b + b b + b b ) b b we hve Usig the lemm for three pirs of umbers b b ) b b ) d b+ b + b b 8 b + b + b b b 8 b + b + b 4

5 b + b + b b b 8 Addig bove iequlities side by side we get b + b + b b + b + b + b + + b b + b b b b b b b ) or b bb b + b + b b with the coditio b + b whe = ) i i i+ i i+ i i+ i= i= i= + Thus b + b + bb b bb b + b b + bb b + b *) i i i i+ i i+ i i i= i= i= i= bi bb b+ bi b i + i= i= bi + bb b bb i i+ bi + bi+ i= i= Accordig to Schur s iequlity we hve Hece from *) we deduce or + bi bb b bi b i + i= i= 4 A ope problem The followig iequlity is problem of USA-TST 000) { } b c Problem 0 Prove tht bc mx b) b c) c ) where bc + + re o-egtive rel umbers 0) The iequlity 0) is beutiful iequlity d there re some wys to prove it eg: see[]) A turl questio rises c the power of 0) be replced by other power? Now we will observe the followig problem Lemm Let f x) = x p x + q where p > 0 x [ b] N the f x) mx{ f f b) } Proof The lemm is proved esily by usig the property of covex fuctio It is left s exercise for the reder Problem Prove tht for ll 0 we hve the followig iequlity { } + mx i i< where N Proof Without losig geerlity we ssume tht So i< { i ) } = ) = + mx Thus ) is equivlet to the followig iequlity ) + + or ) ) + ) 0 Put F ) : = ) ) + ) 5 )

6 It is esy to see tht the form of F ) is the sme s the form of f x) of lemm [ ] i = i i { } ) mx{ F ) F )} Note tht i so usig the lemm for we hve F i i+ i i + Usig the bove property for i = we get F ) mx F t t t ) where t { } i ti { } Suppose tht there exist m umbers tht equl ti i = m ) 0 so m+ m F t t t ) = m + ) m + ) We will prove tht F t t ) 0 Ideed if t m > we put m+ m G ) = F t t t ) = + ) m + G' ) = m + m+ ) + ) m m m ) m = m+ ) + m ) m Sice m d Hece G 0 m ' ) From this we deduce tht G ) G ) = 0 If H m we put ) = F t t t ) = m+ m + ) m + Similrly we lso deduce tht H ) 0 The proof is completed m ) Problem Fid ll rel umbers k such tht we hve the iequlity + b+ c k k bc mx b b c c k for ll o-egtive rel umbers bc { } ) Solutio The swer for this is k = Ideed try b= c= 0 d > from ) we deduce k k l tht Hece for every umber > Let we get k l Whe k = 0) is correct problem 0) We will prove tht ) is icorrect with k > 4 Whe k > try = b= c= where is positive rel umber) we get 4 mx 0 k k = k Hece k This is ot true s teds to ifiity 4 k k = stisfyig ) Therefore there is oly rel umber It is cler tht whe k = the degree of left-hd-side of ) Almost clssicl iequlities re homogeeous iequlities of the first degree 6

7 We coclude this pper with ope problem: for ll o-egtive rel umbers where is turl greter th ) suppose tht there exists fuctio f : = f ) stisfyig the followig three coditios simulteously i) f ) 0 for ll o-egtive rel umbers ii) f ) = 0 if d oly if = = = iii) + f ) for ll rel umbers 0 Prove or disprove tht f ) is homogeeous fuctio of the first degree Refereces [] Ki Y Li Schur s Iequlity Mthemticl Exclibur Vol0 No5 p-4006) Hog Kog [] NVMu Các bài toá ội suy và áp dụg i Vietmese) Eductio Publishig HouseVietm 007 [] PVThu LVi Bất đẳg thức: Suy luậ và khám phá i Vietmese) Vietm tiol uiversity Publishig House Hoi 007 Nguyễ Tiế Lâm studet of K50AS Deprtmet of Mthemtices Mechics Iformtics College of sciece Vietm tiol uiversity Hoi Emil ddress : gtielm@gmilcom 7