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Question-204632

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Question Number 204632 by mr W last updated on 23/Feb/24
Answered by witcher3 last updated on 23/Feb/24
x>1;x=(1/(sin(t)));t∈]0,(π/2)[ ⇒(1/(sin(t)))+((1/(sin(t)))/( (√((1/(sin^2 (t)))−1))))=(1/(sin(t)))+(1/(cos(t)))=2(√2) cauchy shwartz (((1/( (√(sin(t))))))^2 +((1/( (√(cos(t))))))^2 )(((√(sin(t))))^2 +((√(cos^2 (t))))≥(1+1)^2 ⇒(1/(sin(t)))+(1/(cos(t))) ≥(4/( (√2)sin(t+(π/4))))≥(4/( (√2)))=2(√2) equality if only if sin(t)=cos(t)⇒t=(π/4)⇒x=(√2)
$$\left.\mathrm{x}>\mathrm{1};\mathrm{x}=\frac{\mathrm{1}}{\mathrm{sin}\left(\mathrm{t}\right)};\mathrm{t}\in\right]\mathrm{0},\frac{\pi}{\mathrm{2}}\left[\right. \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{sin}\left(\mathrm{t}\right)}+\frac{\frac{\mathrm{1}}{\mathrm{sin}\left(\mathrm{t}\right)}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{t}\right)}−\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{sin}\left(\mathrm{t}\right)}+\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{t}\right)}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{cauchy}\:\mathrm{shwartz}\:\left(\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{sin}\left(\mathrm{t}\right)}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{cos}\left(\mathrm{t}\right)}}\right)^{\mathrm{2}} \right)\left(\left(\sqrt{\mathrm{sin}\left(\mathrm{t}\right)}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{t}\right)}\right)\geqslant\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} \right. \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{sin}\left(\mathrm{t}\right)}+\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{t}\right)}\:\geqslant\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}\mathrm{sin}\left(\mathrm{t}+\frac{\pi}{\mathrm{4}}\right)}\geqslant\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{equality}\:\mathrm{if}\:\mathrm{only}\:\mathrm{if}\: \\ $$$$\mathrm{sin}\left(\mathrm{t}\right)=\mathrm{cos}\left(\mathrm{t}\right)\Rightarrow\mathrm{t}=\frac{\pi}{\mathrm{4}}\Rightarrow\mathrm{x}=\sqrt{\mathrm{2}} \\ $$
Answered by MM42 last updated on 23/Feb/24
let f(x)=x+(x/( (√(x^2 −1)))) & D_f =(1,∞) f′(x)=1−(1/( (x^2 −1)(√(x^2 −1)))) ⇒min_f =((√2),2(√2))
$${let}\:{f}\left({x}\right)={x}+\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\:\:\&\:\:{D}_{{f}} =\left(\mathrm{1},\infty\right) \\ $$$${f}'\left({x}\right)=\mathrm{1}−\frac{\mathrm{1}}{\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\: \\ $$$$\Rightarrow{min}_{{f}} =\left(\sqrt{\mathrm{2}},\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\: \\ $$
Answered by Rasheed.Sindhi last updated on 24/Feb/24
Algebraic Solution 1+(1/( (√(x^2 −1)) ))=((2(√2))/x) ((1/( (√(x^2 −1)) )))^2 =(((2(√2))/x)−1)^2 (1/(x^2 −1))=(8/x^2 )+1−((4(√2))/x) (1/(x^2 −1))=((x^2 −4(√2) x+8)/x^2 ) x^4 −4(√2) x^3 +6x^2 +4(√2) x−8=0 x^2 −4(√2) x+6+((4(√2) )/x)+(1/x^2 )−(9/x^2 )=0 (x^2 +(1/x^2 ))−4(√2) (x−(1/x))+6−(9/x^2 )=0 (x^2 +(1/x^2 )−2)−4(√2) (x−(1/x))+8−(9/x^2 )=0 (x−(1/x))^2 −4(√2) (x−(1/x))+8−(9/x^2 )=0 (x−(1/x))^2 −4(√2) (x−(1/x))+(2(√2) )^2 =(9/x^2 ) (x−(1/x)−2(√2) )^2 =((3/x))^2 x−(1/x)−2(√2) =±(3/x) x^2 −1−2(√2) x=±3 x^2 −2(√2) x−1∓3=0 x^2 −2(√2) x−4=0 ∣ x^2 −2(√2) x+2=0 { ((x=((2(√2) ±(√(8+16)))/2))),((x=((2(√2) ±(√(8−8)))/2))) :} { ((x=((2(√2) ±2(√6))/2)=(√2) ±(√6) (invalid))),((x=((2(√2) )/2)=(√2) ✓ )) :}
$$\boldsymbol{\mathrm{Algebraic}}\:\boldsymbol{\mathrm{Solution}} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{{x}} \\ $$$$\left(\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:}\right)^{\mathrm{2}} =\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}}{{x}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{8}}{{x}^{\mathrm{2}} }+\mathrm{1}−\frac{\mathrm{4}\sqrt{\mathrm{2}}}{{x}} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{{x}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{2}}\:{x}+\mathrm{8}}{{x}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{4}} −\mathrm{4}\sqrt{\mathrm{2}}\:{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{2}}\:{x}−\mathrm{8}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{2}}\:{x}+\mathrm{6}+\frac{\mathrm{4}\sqrt{\mathrm{2}}\:}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{9}}{{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)−\mathrm{4}\sqrt{\mathrm{2}}\:\left({x}−\frac{\mathrm{1}}{{x}}\right)+\mathrm{6}−\frac{\mathrm{9}}{{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2}\right)−\mathrm{4}\sqrt{\mathrm{2}}\:\left({x}−\frac{\mathrm{1}}{{x}}\right)+\mathrm{8}−\frac{\mathrm{9}}{{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{2}}\:\left({x}−\frac{\mathrm{1}}{{x}}\right)+\mathrm{8}−\frac{\mathrm{9}}{{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{2}}\:\left({x}−\frac{\mathrm{1}}{{x}}\right)+\left(\mathrm{2}\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} =\frac{\mathrm{9}}{{x}^{\mathrm{2}} } \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}−\mathrm{2}\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} =\left(\frac{\mathrm{3}}{{x}}\right)^{\mathrm{2}} \\ $$$${x}−\frac{\mathrm{1}}{{x}}−\mathrm{2}\sqrt{\mathrm{2}}\:=\pm\frac{\mathrm{3}}{{x}} \\ $$$${x}^{\mathrm{2}} −\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}}\:{x}=\pm\mathrm{3} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}\:{x}−\mathrm{1}\mp\mathrm{3}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}\:{x}−\mathrm{4}=\mathrm{0}\:\mid\:{x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}\:{x}+\mathrm{2}=\mathrm{0} \\ $$$$\begin{cases}{{x}=\frac{\mathrm{2}\sqrt{\mathrm{2}}\:\pm\sqrt{\mathrm{8}+\mathrm{16}}}{\mathrm{2}}}\\{{x}=\frac{\mathrm{2}\sqrt{\mathrm{2}}\:\pm\sqrt{\mathrm{8}−\mathrm{8}}}{\mathrm{2}}}\end{cases}\: \\ $$$$\begin{cases}{{x}=\frac{\mathrm{2}\sqrt{\mathrm{2}}\:\pm\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{2}}=\sqrt{\mathrm{2}}\:\pm\sqrt{\mathrm{6}}\:\left({invalid}\right)}\\{{x}=\frac{\mathrm{2}\sqrt{\mathrm{2}}\:}{\mathrm{2}}=\sqrt{\mathrm{2}}\:\checkmark\:}\end{cases}\: \\ $$

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