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Question Number 39971 by math2018 last updated on 14/Jul/18

$$a>0,b>0,$$ $$Whatistheminimumvalueof$$ $$\frac{b^2+2}{a+b}+\frac{a^2}{ab+1}?$$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18

$$\frac{b^2}{a+b}+\frac{a^2}{ab+1}+\frac{2}{a+b}=x\left(say\right)$$ $$usingAM>GM$$ $$\frac{a+b}{2}>\sqrt{ab}$$ $$$$ $$\frac{\frac{b^2}{2}}{\frac{a+b}{2}}+\frac{\frac{a^2}{2}}{\frac{ab+1}{2}}+\frac{\frac{2}{2}}{\frac{a+b}{2}}<\frac{\frac{b^2}{2}}{\sqrt{ab}}+\frac{\frac{a^2}{2}}{\sqrt{ab}}+\frac{\frac{2}{2}}{\sqrt{ab}}\to y\left(say\right)$$ $$x<\frac{b^2}{2\sqrt{ab}}+\frac{a^2}{2\sqrt{ab}}+\frac{2}{2\sqrt{ab}}$$ $$x<\frac{a^2+b^2+2}{2\sqrt{ab}}$$ $$x<\frac{\frac{a^2+b^2}{2}\times 2+2}{2\sqrt{ab}}$$ $$using\frac{a^2+b^2}{2}>\sqrt{a^2{b}^2}$$ $$\frac{\frac{a^2+b^2}{2}\times 2+2}{2\sqrt{ab}}>\frac{2ab+2}{2\sqrt{ab}}\to z\left(say\right)$$ $$y>\frac{ab+1}{\sqrt{ab}}$$ $$y>\frac{\frac{ab+1}{2}\times 2}{\sqrt{ab}}\to z$$ $$$$ $$\frac{\frac{ab+1}{2}\times 2}{\sqrt{ab}}>\frac{\sqrt{ab}\times 2}{\sqrt{ab}}$$ $$soz>2$$ $$givenexpressionassumedtobex$$ $$x<y$$ $$y>z$$ $$z>2$$ $$icannotcorelatebetweenxandz$$ $$$$ $$$$ $$$$

Commented bymath2018 last updated on 15/Jul/18

$$Thankyouverymuch!$$

Answered by ajfour last updated on 14/Jul/18

$$f(a,b)=\frac{b^2+2}{a+b}+\frac{a^2}{ab+1}$$ $$<\frac{b^2+2}{2\sqrt{ab}}+\frac{a^2}{2\sqrt{ab}}=\frac{{(a-b)}^2+2ab+2}{2\sqrt{ab}}$$ $$=\frac{{(a-b)}^2}{2\sqrt{ab}}+\sqrt{ab}+\frac{1}{\sqrt{ab}}$$ $$theaboveexpressionhasa$$ $$minimumofvalueequalto2$$ $$fora=b=1$$ $$hence{f}_{min}(a,b)⩽2.$$ $$$$

Commented bymath2018 last updated on 15/Jul/18

$$ThankyouSir!$$ $$$$

Answered by math2018 last updated on 15/Jul/18

$$1°a+b⩾ab+1$$ $$\frac{b^2+2}{a+b}+\frac{a^2}{ab+1}⩾\frac{(a^2+1)+(b^2+1)}{a+b}⩾\frac{2a+2b}{a+b}=2;$$ $$2°a+b<ab+1$$ $$\frac{b^2+2}{a+b}+\frac{a^2}{ab+1}>\frac{a^2+b^2+2}{ab+1}⩾\frac{2ab+2}{ab+1}=2$$ $$hencetheminimumvalueis2.$$

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