Question Number 115033 by bobhans last updated on 23/Sep/20

If a and b positive real number where  a^(505)  + b^(505)  = 1, then minimum value  a^(2020)  + b^(2020)  is __

Answered by 1549442205PVT last updated on 23/Sep/20

Put a^(505) =x,b^(505) =y⇒x+y=1.We need  find minimum value of P=x^4 +y^4   Since x,y>0,1=x+y=((√x)−(√y))^2 +2(√(xy))≥2(√(xy))  ⇒(√(xy))≤1/2⇒xy≤1/4(1).Hence,  x^4 +y^4 =(x+y)^4 −4xy(x^2 +y^4 )−6x^2 y^2   =1−4xy[(x+y)^2 −2xy]−6x^2 y^2   =1−4xy(1−2xy)−6(xy)^2 =1−4xy+2(xy)^2   =2(xy−(1/4))^2 −3xy+(7/8)≥0−3.(1/4)+(7/8)  =(1/8).The equality ocurrs if and only  if x=y=1/2⇔a=b= ^(505) (√(1/2))  Thus (a^(2020) +b^(2020) )_(min) =(1/8)  when a=b=(1/(  ^(505) (√2)))

Answered by bemath last updated on 23/Sep/20