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Question Number 110704 by Aina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{Find}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}$$$$\frac{{\mathrm{n}}^2+1}{\mathrm{n}}+\frac{\mathrm{n}}{{\mathrm{n}}^2+1},\mathrm{n}>0$$

Answered by mr W last updated on 30/Aug/20

$$\frac{n^2+1}{n}+\frac{n}{n^2+1}$$$$=(n+\frac{1}{n})+\frac{1}{(n+\frac{1}{n})}$$$$n+\frac{1}{n}⩾2\sqrt{n\times \frac{1}{n}}=2$$$$sothequestionistofindtheminimum$$$$offunctionf\left(x\right)=x+\frac{1}{x}forx⩾2.$$$$f^{\prime }\left(x\right)=1-\frac{1}{x^2}>0forx>1$$$$thatmeansf\left(x\right)isstrictlyincreasing$$$$forx>1,soforx⩾2wehave$$$$f{\left(x\right)}_{min}=f\left(2\right)=2+\frac{1}{2}=\frac{5}{2}.$$

Answered by floor(10²Eta[1]) last updated on 30/Aug/20

$$\mathrm{with}\mathrm{AM}-\mathrm{GM}\mathrm{inequality}:$$$$\frac{\frac{{\mathrm{n}}^2+1}{\mathrm{n}}+\frac{\mathrm{n}}{{\mathrm{n}}^2+1}}{2}⩾\sqrt{\left(\frac{{\mathrm{n}}^2+1}{\mathrm{n}}\right)\left(\frac{\mathrm{n}}{{\mathrm{n}}^2+1}\right)}=1$$$$\Rightarrow \frac{{\mathrm{n}}^2+1}{\mathrm{n}}+\frac{\mathrm{n}}{{\mathrm{n}}^2+1}⩾2$$$$\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{is}2.$$

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{Thanks}\mathrm{but}\mathrm{the}\mathrm{answer}\mathrm{is}5/2.\mathrm{I}\mathrm{was}$$$$\mathrm{also}\mathrm{surprised}\mathrm{when}\mathrm{I}\mathrm{found}\mathrm{out}\mathrm{it}\mathrm{was}$$$$5/2.$$

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{why}\mathrm{AM}-\mathrm{GM}{\mathrm{doesn}}^{\prime }\mathrm{t}$$$$\mathrm{work}\mathrm{here}.$$

Commented by mnjuly1970 last updated on 30/Aug/20

$$if(min=2)then:$$$$$$$$\frac{n^2+1}{n}=\frac{n}{n^2+1}\Rightarrow n^4+2n^2+1=n^2$$$$n^4+n^2+1=0\Rightarrow n\notin \mathbb{R}$$$$$$$$solution:\mathrm{F}\left(n\right)=\left(n\stackrel{\left[min\right]]}{+}\frac{1}{n}\right)+\left(\frac{\stackrel{\left[max\right]}{1}}{n+\frac{1}{n}}\right)$$$$min\left(\mathrm{F}\left(n\right)\right)=2+\frac{1}{2}=\frac{5}{2}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{Thanks}.$$

Answered by $@y@m last updated on 30/Aug/20

$$Lety=\frac{{\mathrm{n}}^2+1}{\mathrm{n}}+\frac{\mathrm{n}}{{\mathrm{n}}^2+1},\mathrm{n}>0$$$$\Rightarrow y=\frac{{(n^2+1)}^2+n^2}{n(n^2+1)}$$$$\Rightarrow y=\frac{n^4+3n^2+1}{n^3+n}$$$$\frac{dy}{dn}=\frac{(n^3+n)(4n^3+6n)-(n^4+3n^2+1)(3n^2+1)}{{(n^3+n)}^2}$$$$\frac{dy}{dn}=\frac{4n^6+10n^4+6n^2-(3n^6+10n^4+6n^2+1)}{{(n^3+n)}^2}$$$$\frac{dy}{dn}=\frac{n^6-1}{{(n^3+n)}^2}....\left(1\right)$$$$Formaximaorminima,$$$$\frac{dy}{dn}=0$$$$\Rightarrow n^6-1=0$$$$\Rightarrow n=1$$$$Now,\frac{d^{2}y}{{dn}^2}=\frac{{(n^3+n)}^{2}6n^5-(n^6-1).2(n^3+n)(3n^2+1)}{{(n^3+n)}^4}$$$${\left[\frac{d^{2}y}{{dn}_{}^2}\right]}_{atn=1}=\frac{4.6.1-0}{2^4}=\frac{3}{2}>0$$$$\therefore yisminimumatn=1$$$$\Rightarrow y_{min}=\frac{1^2+1}{1}+\frac{1}{1^2+1}=2+\frac{1}{2}=\frac{5}{2}$$$$$$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

$$\mathrm{Thanks}.$$

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