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Question Number 114181 by bemath last updated on 17/Sep/20

$$Givenafunction$$ $$f\left(x\right)=x^2+\frac{1}{x^2}+4x+\frac{4}{x};wherex>0.$$ $$findtheminimumvalueoff\left(x\right)$$

Answered by bobhans last updated on 17/Sep/20

$$recallp+\frac{1}{p}⩾2;p>0$$ $$nowf\left(x\right)=(x^2+\frac{1}{x^2})+4(x+\frac{1}{x})$$ $$\Rightarrow f\left(x\right)={(x+\frac{1}{x})}^2-2+4(x+\frac{1}{x})$$ $$f\left(x\right){\mid }_{minimum}=2^2-2+4.2=10.$$

Answered by mathmax by abdo last updated on 17/Sep/20

$${\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}⩾2\mathrm{and}\mathrm{x}+\frac{1}{\mathrm{x}}⩾2\Rightarrow 4(\mathrm{x}+\frac{1}{\mathrm{x}})⩾8\Rightarrow {\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}+4(\mathrm{x}+\frac{1}{\mathrm{x}})⩾10\Rightarrow$$ $${\min }_{{\mathrm{R}}^{+}}\mathrm{f}\left(\mathrm{x}\right)=10$$

Answered by 1549442205PVT last updated on 18/Sep/20

$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}+4(\mathrm{x}+\frac{1}{\mathrm{x}})$$ $$={(\mathrm{x}-\frac{1}{\mathrm{x}})}^2+4{(\sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}})}^2+10⩾10$$ $$\mathrm{The}\mathrm{equality}\mathrm{ocurrs}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}$$ $$\mathrm{x}=1.\mathrm{Hence},\mathrm{f}{\left(\mathrm{x}\right)}_{\min }=10\mathrm{when}\mathrm{x}=2$$

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