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Question Number 190192 by mr W last updated on 29/Mar/23

$$ifa>b>0,findtheminimumof$$ $$a^2+\frac{1}{(a-b)b}=?$$

Commented byFrix last updated on 29/Mar/23

$$\mathrm{Minimum}=4\mathrm{at}a=\sqrt{2}\wedge b=\frac{\sqrt{2}}{2}$$

Commented bymr W last updated on 29/Mar/23

$${what}^{\prime }syoursolution?$$

Answered by cortano12 last updated on 29/Mar/23

$$\mathrm{z}={\mathrm{a}}^2+{(\mathrm{ab}-{\mathrm{b}}^2)}^{-1}$$ $$\frac{\mathrm{dz}}{\mathrm{da}}=2\mathrm{a}-\mathrm{b}{(\mathrm{ab}-{\mathrm{b}}^2)}^{-2}$$ $$\frac{\mathrm{dz}}{\mathrm{db}}=-(\mathrm{a}-2\mathrm{b}){(\mathrm{ab}-{\mathrm{b}}^2)}^{-2}$$ $$\frac{\mathrm{dz}}{\mathrm{da}}=0\Rightarrow 2\mathrm{a}=\frac{\mathrm{b}}{{\mathrm{b}}^2{(\mathrm{a}-\mathrm{b})}^2}$$ $$2\mathrm{a}=\frac{1}{\mathrm{b}{(\mathrm{a}-\mathrm{b})}^2}\left(\mathrm{i}\right)$$ $$\frac{\mathrm{dz}}{\mathrm{db}}=0\Rightarrow \frac{2\mathrm{b}-\mathrm{a}}{{(\mathrm{ab}-{\mathrm{b}}^2)}^2}=0;\mathrm{a}=2\mathrm{b}\left(\mathrm{ii}\right)$$ $$\Rightarrow 4\mathrm{b}=\frac{1}{{\mathrm{b}}^3};\mathrm{b}=\frac{1}{\sqrt{2}}\wedge \mathrm{a}=\sqrt{2}$$ $$\Rightarrow {\mathrm{z}}_{\min }=2+\frac{\sqrt{2}}{(\sqrt{2}-\frac{1}{2}\sqrt{2})}=4$$

Commented bymr W last updated on 29/Mar/23

$$thanks!$$

Answered by witcher3 last updated on 29/Mar/23

$$(\mathrm{a}-\mathrm{b})\mathrm{b}⩽\frac{1}{4}{((\mathrm{a}-\mathrm{b})+\mathrm{b})}^2...\mathrm{AM},\mathrm{GM}$$ $$\Rightarrow \frac{1}{\mathrm{b}(\mathrm{a}-\mathrm{b})}⩾\frac{4}{{\mathrm{a}}^2}$$ $${\mathrm{a}}^2+\frac{1}{(\mathrm{a}-\mathrm{b})\mathrm{b}}⩾{\mathrm{a}}^2+\frac{4}{{\mathrm{a}}^2}⩾2\sqrt{.{\mathrm{a}}^2.\frac{4}{{\mathrm{a}}^2}}=4...\mathrm{AM},\mathrm{GM}$$

Commented bymr W last updated on 29/Mar/23

$$thanks!$$

Answered by mr W last updated on 29/Mar/23

$$sayc=a-b>0$$ $$a^2+\frac{1}{(a-b)b}$$ $$={(c+b)}^2+\frac{1}{cb}$$ $$=c^2+b^2+2bc+\frac{1}{cb}$$ $$=c^2+b^2+bc+bc+\frac{1}{4cb}+\frac{1}{4cb}+\frac{1}{4cb}+\frac{1}{4cb}$$ $$⩾8\sqrt[8]{c^2\times b^2\times {\left(bc\right)}^2\times {\left(\frac{1}{4cb}\right)}^4}=8\times \frac{1}{2}=4$$ $$\Rightarrow minimum=4$$ $$when{c}^2=b^2=bc=\frac{1}{4cb},i.e.c=b=\frac{1}{\sqrt{2}}\mathrm{\&}a=\sqrt{2}$$

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