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Question Number 99645 by bramlex last updated on 22/Jun/20

If x,y > 0 then ∣(√(xy))−((x+y)/2)∣ + ∣((x+y)/2) + (√(xy)) ∣ =

Ifx,y>0thenxyx+y2+x+y2+xy=

Commented bybemath last updated on 22/Jun/20

since (√(xy)) > 0 & ((x+y)/2) > 0  so ((x+y)/2) + (√(xy)) > 0  case(1) if (√(xy))−((x+y)/2) > 0  then (√(xy)) −((x+y)/2) + ((x+y)/2) +(√(xy)) = 2(√(xy ))  case(2) if (√(xy))−((x+y)/2) < 0  then −(√(xy))+((x+y)/2) + ((x+y)/2)+(√(xy)) = x+y

sincexy>0&x+y2>0 sox+y2+xy>0 case(1)ifxyx+y2>0 thenxyx+y2+x+y2+xy=2xy case(2)ifxyx+y2<0 thenxy+x+y2+x+y2+xy=x+y

Answered by Farruxjano last updated on 22/Jun/20

When x,y>0 ⇒ ((x+y)/2) + (√(xy))>0  ⇒  ∣((x+y)/2) + (√(xy)) ∣=((x+y)/2) + (√(xy))  We know inequality Koshi:  ((x+y)/2)≥(√(xy))   when  x,y>0  So,  ∣(√(xy))−((x+y)/2)∣=((x+y)/2)−(√(xy))    ∣(√(xy))−((x+y)/2)∣ + ∣((x+y)/2) + (√(xy)) ∣ =((x+y)/2)−(√(xy))+  +((x+y)/2) + (√(xy))=x+y ▲.

Whenx,y>0x+y2+xy>0 x+y2+xy∣=x+y2+xy WeknowinequalityKoshi: x+y2xywhenx,y>0So, xyx+y2∣=x+y2xy xyx+y2+x+y2+xy=x+y2xy+ +x+y2+xy=x+y.

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