Question Number 115058 by mnjuly1970 last updated on 23/Sep/20

       ....  nice  calculus ...        a , b , c , d  ∈N and                                   (1/a)+(1/b)+(1/c)+(1/d)=(1/2)                            find                                                 max(a+b+c+d) =???        ...m.n.july.1970...

Answered by 1549442205PVT last updated on 25/Sep/20

  (1/a)+(1/b)+(1/c)+(1/d)=(1/2)(1)  Since a,b,c,d are equal in role WLOG  we can suppose that a≥b≥c≥d  ⇒(4/a)≤  (1/a)+(1/b)+(1/c)+(1/d)=(1/2)≤(4/d)  ⇒a≥8≥d≥3  i)Case d=3⇒  (1/a)+(1/b)+(1/c)=(1/2)−(1/3)=(1/6)  ⇒(3/a)≤  (1/a)+(1/b)+(1/c)=(1/6)≤(3/c)  ⇒a≥18≥c≥7  a)For c=7⇒  (1/a)+(1/b)=(1/6)−(1/7)=(1/(42))  ⇒(2/a)≤(1/a)+(1/b)=(1/(42))≤(2/b)  ⇒a≥84≥b≥43  •if b=43 then (1/a)=(1/(42))−(1/(43))=(1/(1806))  ⇒a=1806  Thus,a+b+c+d=1806+43+7+3=1859  •If b=44⇒(1/a)=(1/(42))−(1/(44))=(1/(924))⇒a=924  42(a+b)=ab⇒(a−42)(b−42)=1764  =4.21^2 =2^2 .3^2 .7^2   b)For c=8⇒(1/a)+(1/b)=(1/6)−(1/8)=(1/(24))  ⇒ab=24(a+b)⇔(a−24)(b−24)=576  =24^2 =2^6 .3^2   b=25⇒a=600⇒a+b+c+d=636<1859  c)For c=9,10,...,18 by  similar argument  we see that a+b+c+d<1859  ii)d=4⇒(1/a)+(1/b)+(1/c)=(1/2)−(1/4)=(1/4)  ⇒(3/a) ≤(1/4)≤(3/c)⇒a≥12≥c≥5  a)For c=5⇒(1/a)+(1/b)=(1/4)−(1/5)=(1/(20))  ⇒ab=20(a+b)⇔(a−20)(b−20)=400=2^4 .5^2   •if b=21⇒a=420⇒a+b+c+d=450<1859  b)For c=6...12^(−) ,by similar argument  we get a+b+c+d<1859  ii)Cases d=5...8^(−) ,repeating above argument   we obtain a+b+c+d<1859  Thus,if   (1/a)+(1/b)+(1/c)+(1/d)=(1/2)  with a,b,c,d∈N then  max(a+b+c+d)=1859

Commented bymnjuly1970 last updated on 23/Sep/20

verh excellent. thank you..