Question Number 71206 by naka3546 last updated on 13/Oct/19

Let p,q,r are positive real numbers . 0 < r < min{p,q}. Prove that (√(p−r)) + (√(q−r)) ≤ min{(√((pq)/r)) , (√(2(p+q − 2r))) }

Answered by mind is power last updated on 13/Oct/19

(√(p−r ))+(√(q−r ))=≤(√(2(p+q−2r))) (√x)+(√y)≤(√(2x+2y)) cause x+y−2(√(xy))=((√x)−(√y))^2 ≥0 ⇒x+y+2(√(xy))≤2x+2y ⇒(√x)+(√y)≤(√(2(x+y))) ⇒x=p−r y=q−r (√(p−r))+(√(q−r))≤(√(2(p+q−2r)))....1 (√(p−r))+(√(q−r))≤(√((pq)/r)) (√(p−r))+(√(q−r))=(√r)(√((p/r)−1))+(√r)(√((q/r)−1)) (√(x−1))+(√(y−1))≤(√(xy )) x+y−2+2(√((x−1)(y−1)))≤xy 2(√((x−1)(y−1)))≤xy−x−y+2=(x−1)(y−1)+1 ⇒((√((x−1)(y−1)))−1)^2 ≥0 True x=(p/r),y=(q/r)⇒ (√((p/r)−1))+(√((q/r)−1))≤(√((pq)/r^2 )) ⇒(√r)((√((p/r)−1))+(√((q/r)−1)))≤(√((pq)/r))...2 2&1⇒ (√(p−r))+(√(q−r))≤min{(√((pq)/r)),(√(2(p+q−2r)))}