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Question Number 152063 by Tawa11 last updated on 25/Aug/21

$$\mathrm{If}\mathrm{x}\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{number}\mathrm{and}\mathrm{y}\mathrm{is}\mathrm{equal}\mathrm{to}\frac{{\mathrm{x}}^2+1}{{\mathrm{x}}^2+\mathrm{x}+1},$$$$\mathrm{show}\mathrm{that}\mid \mathrm{y}-\frac{4}{3}\mid ⩽\frac{2}{3}$$

Answered by mr W last updated on 25/Aug/21

$$y=\frac{x^2+1}{x^2+x+1}$$$$y=1-\frac{x}{x^2+x+1}$$$$y=1-\frac{1}{x+\frac{1}{x}+1}$$$$x+\frac{1}{x}⩾2forx>0$$$$x+\frac{1}{x}⩽-2forx<0$$$$y_{max}=1-\frac{1}{-2+1}=2$$$$y_{min}=1-\frac{1}{2+1}=\frac{2}{3}$$$$\Rightarrow \frac{2}{3}⩽y⩽2$$$$\Rightarrow -\frac{2}{3}⩽y-\frac{4}{3}⩽\frac{2}{3}$$$$\Rightarrow \mid y-\frac{4}{3}\mid ⩽\frac{2}{3}$$

Commented by Tawa11 last updated on 25/Aug/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Commented by otchereabdullai@gmail.com last updated on 25/Aug/21

$$\mathrm{wow}!$$

Answered by john_santu last updated on 25/Aug/21

$$\mathrm{y}=\frac{{\mathrm{x}}^2+1}{{\mathrm{x}}^2+\mathrm{x}+1}\Rightarrow {\mathrm{yx}}^2+\mathrm{yx}+\mathrm{y}={\mathrm{x}}^2+1$$$$\Rightarrow (\mathrm{y}-1){\mathrm{x}}^2+\mathrm{yx}+\mathrm{y}-1=0$$$$\mathrm{\Delta }⩾0\Rightarrow {\mathrm{y}}^2-4{(\mathrm{y}-1)}^2⩾0$$$$\Rightarrow {\mathrm{y}}^2-{(2\mathrm{y}-2)}^2⩾0$$$$\Rightarrow (3\mathrm{y}-2)(-\mathrm{y}+2)⩾0$$$$\Rightarrow (3\mathrm{y}-2)(\mathrm{y}-2)⩽0$$$$\Rightarrow \frac{2}{3}⩽\mathrm{y}⩽2;\frac{2}{3}-\frac{4}{3}⩽\mathrm{y}-\frac{4}{3}⩽2-\frac{4}{3}$$$$\Rightarrow -\frac{2}{3}⩽\mathrm{y}-\frac{4}{3}⩽\frac{2}{3}$$$$\Rightarrow \mid \mathrm{y}-\frac{4}{3}\mid ⩽\frac{2}{3}.$$

Commented by Tawa11 last updated on 25/Aug/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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