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Question Number 115033 by bobhans last updated on 23/Sep/20

$$Ifaandbpositiverealnumberwhere$$$$a^{505}+b^{505}=1,thenminimumvalue$$$$a^{2020}+b^{2020}is\mathrm{\_}\mathrm{\_}$$

Answered by 1549442205PVT last updated on 23/Sep/20

$$\mathrm{Put}{\mathrm{a}}^{505}=\mathrm{x},{\mathrm{b}}^{505}=\mathrm{y}\Rightarrow \mathrm{x}+\mathrm{y}=1.\mathrm{We}\mathrm{need}$$$$\mathrm{find}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{P}={\mathrm{x}}^4+{\mathrm{y}}^4$$$$\mathrm{Since}\mathrm{x},\mathrm{y}>0,1=\mathrm{x}+\mathrm{y}={(\sqrt{\mathrm{x}}-\sqrt{\mathrm{y}})}^2+2\sqrt{\mathrm{xy}}⩾2\sqrt{\mathrm{xy}}$$$$\Rightarrow \sqrt{\mathrm{xy}}⩽1/2\Rightarrow \mathrm{xy}⩽1/4\left(1\right).\mathrm{Hence},$$$${\mathrm{x}}^4+{\mathrm{y}}^4={(\mathrm{x}+\mathrm{y})}^4-4\mathrm{xy}({\mathrm{x}}^2+{\mathrm{y}}^4)-{6\mathrm{x}}^2{\mathrm{y}}^2$$$$=1-4\mathrm{xy}[{(\mathrm{x}+\mathrm{y})}^2-2\mathrm{xy}]-{6\mathrm{x}}^2{\mathrm{y}}^2$$$$=1-4\mathrm{xy}(1-2\mathrm{xy})-6{\left(\mathrm{xy}\right)}^2=1-4\mathrm{xy}+2{\left(\mathrm{xy}\right)}^2$$$$=2{(\mathrm{xy}-\frac{1}{4})}^2-3\mathrm{xy}+\frac{7}{8}⩾0-3.\frac{1}{4}+\frac{7}{8}$$$$=\frac{1}{8}.\mathrm{The}\mathrm{equality}\mathrm{ocurrs}\mathrm{if}\mathrm{and}\mathrm{only}$$$$\mathrm{if}\mathrm{x}=\mathrm{y}=1/2\iff \mathrm{a}=\mathrm{b}={}^{505}\sqrt{\frac{1}{2}}$$$$\mathrm{Thus}{({\mathrm{a}}^{2020}+{\mathrm{b}}^{2020})}_{\min }=\frac{1}{8}$$$$\mathrm{when}\mathrm{a}=\mathrm{b}=\frac{1}{{}^{505}\sqrt{2}}$$

Answered by bemath last updated on 23/Sep/20

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