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Question Number 151609 by mathdanisur last updated on 22/Aug/21
Answered by ghimisi last updated on 22/Aug/21
⇔logxy(1+xy)2⩾logx+y2(x+y2+1)⇔⇔ln(1+xy)lnxy⩾ln(1+x+y2)lnx+y2(∙)f(t)=ln(1+t)lnt,f:(1;∞)→Rf′(t)=lnt1+t−ln(1+t)tln2t=tlnt−(t+1)ln(t+1)t(t+1)ln2t<0f↘;xy⩽x+y2⇒f(xy)⩾f(x+y2)⇒(∙)
Commented by mathdanisur last updated on 22/Aug/21
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