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Question Number 198147 by mr W last updated on 11/Oct/23
ifa,x,y,bisanAPanda,p,q,bisaGP.provethatxy⩾pq.(witha,b>0)
Answered by deleteduser1 last updated on 11/Oct/23
p2=aq;q2=pb⇒p2q2=abpq⇒pq=ab2x=a+y;2y=x+b⇒4xy=(a+y)(x+b)3xy=ax+ab+yb⩾?3pq=3ab⇔ax+yb⩾2ab⇔a(a+d)+(a+2d)(a+3d)⩾2(a)(a+3d)a2+ad+a2+3ad+2ad+6d2⩾2a2+6ad⇔d2⩾0(true)..Hence3xy⩾3ab⇒xy⩾ab=pqEqualityholdswhend=0⇒a=b=x=y=p=q
Commented by mr W last updated on 11/Oct/23
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