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Question Number 80039 by jagoll last updated on 30/Jan/20

$$provethat$$$$(1+x)(1+\frac{1}{x})⩾4$$

Commented by mr W last updated on 30/Jan/20

$$onlyforx\in R^{+}$$$$1+x⩾2\sqrt{x}$$$$1+\frac{1}{x}⩾2\frac{1}{\sqrt{x}}$$$$(1+x)(1+\frac{1}{x})⩾2\sqrt{x}\times 2\frac{1}{\sqrt{x}}=4$$$$or$$$$(1+x)(1+\frac{1}{x})=2+x+\frac{1}{x}⩾2+2=4$$$$$$$$forx\in R^{-}$$$$(1+x)(1+\frac{1}{x})=2+x+\frac{1}{x}=2-(-x+\frac{1}{-x})⩽2-2=0$$

Commented by Tony Lin last updated on 30/Jan/20

$$[1^2+{\left(\sqrt{x}\right)}^2][1^2+{\left(\frac{1}{\sqrt{x}}\right)}^2]⩾{(1+1)}^2=4$$

Commented by MJS last updated on 30/Jan/20

$$y=(1+x)(1+\frac{1}{x})$$$$\mathrm{defined}\mathrm{for}x\in \mathbb{R}\mathrm{\setminus }\left\{0\right\}$$$$(1+x)(1+\frac{1}{x})<4$$$$x+\frac{1}{x}<2$$$$\left(1\right)x>0$$$$x^2-2x+1<0$$$${(x-1)}^2<0\mathrm{wrong}\Rightarrow (1+x)(1+\frac{1}{x})⩾4$$$$\left(2\right)x<0$$$$x^2-2x+1>0$$$${(x-1)}^2>0\mathrm{true}\mathrm{\forall }x<0\Rightarrow (1+x)(1+\frac{1}{x})<4$$$$\{\begin{array}{l}(1+x)(1+\frac{1}{x})⩾4;x>0\\ (1+x)(1+\frac{1}{x})<4;x<0\end{array}$$

Commented by mr W last updated on 30/Jan/20

$$typosir?$$$$(1+x)(1+\frac{1}{x})<4;x<0$$$$shouldbe:$$$$(1+x)(1+\frac{1}{x})⩽0;x<0$$

Commented by MJS last updated on 30/Jan/20

$$\mathrm{You}\mathrm{are}\mathrm{right}\mathrm{but}\mathrm{with}\mathrm{my}\mathrm{method}\mathrm{I}\mathrm{only}$$$$\mathrm{examined}\mathrm{if}{\mathrm{it}}^{\prime }\mathrm{s}<4.$$

Commented by jagoll last updated on 31/Jan/20

$$thankyou$$

Answered by $@ty@m123 last updated on 31/Jan/20

$$Lety=(1+x)(1+\frac{1}{x})$$$$\Rightarrow y=2+x+\frac{1}{x}$$$$\frac{dy}{dx}=1-\frac{1}{x^2}$$$$Formax.ormin.$$$$\frac{dy}{dx}=0$$$$1-\frac{1}{x^2}=0$$$$\Rightarrow x=\pm 1$$$$y{\mid }_{x=1}=4,$$$$y{\mid }_{x=-1}=0$$$$\frac{d^{2}y}{{dx}^2}=\frac{1}{x^3}$$$${\left[\frac{d^{2}y}{{dx}^2}\right]}_{x=1}=1>0$$$$\therefore yismininmumatx=1$$$$andmin.\left(y\right)=4$$

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