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

If  x  is a real number and   y  is equal to     ((x^2   +  1)/(x^2   +  x  +  1)),  show that     ∣y   −   (4/3)∣   ≤   (2/3)

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

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

y=((x^2 +1)/(x^2 +x+1))  y=1−(x/(x^2 +x+1))  y=1−(1/(x+(1/x)+1))  x+(1/x)≥2 for x>0  x+(1/x)≤−2 for x<0  y_(max) =1−(1/(−2+1))=2  y_(min) =1−(1/(2+1))=(2/3)  ⇒(2/3)≤y≤2  ⇒−(2/3)≤y−(4/3)≤(2/3)  ⇒∣y−(4/3)∣≤(2/3)

$${y}=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$${y}=\mathrm{1}−\frac{{x}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$${y}=\mathrm{1}−\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}} \\ $$$${x}+\frac{\mathrm{1}}{{x}}\geqslant\mathrm{2}\:{for}\:{x}>\mathrm{0} \\ $$$${x}+\frac{\mathrm{1}}{{x}}\leqslant−\mathrm{2}\:{for}\:{x}<\mathrm{0} \\ $$$${y}_{{max}} =\mathrm{1}−\frac{\mathrm{1}}{−\mathrm{2}+\mathrm{1}}=\mathrm{2} \\ $$$${y}_{{min}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}+\mathrm{1}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{\mathrm{2}}{\mathrm{3}}\leqslant{y}\leqslant\mathrm{2} \\ $$$$\Rightarrow−\frac{\mathrm{2}}{\mathrm{3}}\leqslant{y}−\frac{\mathrm{4}}{\mathrm{3}}\leqslant\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\mid{y}−\frac{\mathrm{4}}{\mathrm{3}}\mid\leqslant\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Commented by Tawa11 last updated on 25/Aug/21

Thanks sir. God bless you.

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

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

wow!

$$\mathrm{wow}! \\ $$

Answered by john_santu last updated on 25/Aug/21

y=((x^2 +1)/(x^2 +x+1)) ⇒yx^2 +yx+y=x^2 +1  ⇒(y−1)x^2 +yx+y−1 = 0  Δ≥0 ⇒y^2 −4(y−1)^2 ≥0  ⇒y^2 −(2y−2)^2 ≥0  ⇒(3y−2)(−y+2)≥0  ⇒(3y−2)(y−2)≤0  ⇒(2/3)≤y≤2 ; (2/3)−(4/3)≤y−(4/3)≤2−(4/3)  ⇒−(2/3)≤y−(4/3)≤(2/3)  ⇒ ∣y−(4/3)∣≤(2/3) .

$$\mathrm{y}=\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:\Rightarrow\mathrm{yx}^{\mathrm{2}} +\mathrm{yx}+\mathrm{y}=\mathrm{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{y}−\mathrm{1}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{yx}+\mathrm{y}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\Delta\geqslant\mathrm{0}\:\Rightarrow\mathrm{y}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{y}^{\mathrm{2}} −\left(\mathrm{2y}−\mathrm{2}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3y}−\mathrm{2}\right)\left(−\mathrm{y}+\mathrm{2}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3y}−\mathrm{2}\right)\left(\mathrm{y}−\mathrm{2}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{2}}{\mathrm{3}}\leqslant\mathrm{y}\leqslant\mathrm{2}\:;\:\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{3}}\leqslant\mathrm{y}−\frac{\mathrm{4}}{\mathrm{3}}\leqslant\mathrm{2}−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow−\frac{\mathrm{2}}{\mathrm{3}}\leqslant\mathrm{y}−\frac{\mathrm{4}}{\mathrm{3}}\leqslant\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\:\mid\mathrm{y}−\frac{\mathrm{4}}{\mathrm{3}}\mid\leqslant\frac{\mathrm{2}}{\mathrm{3}}\:. \\ $$

Commented by Tawa11 last updated on 25/Aug/21

Thanks sir. God bless you.

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

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