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Question Number 158245 by HongKing last updated on 01/Nov/21
Prove that: (((x-1)^2 )/x) + ((x+1)/( (√(x^2 +1)))) ≥ (√2) ; ∀x>0
$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\left(\mathrm{x}-\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{x}}\:+\:\frac{\mathrm{x}+\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\:\geqslant\:\sqrt{\mathrm{2}}\:\:;\:\:\forall\mathrm{x}>\mathrm{0} \\ $$$$ \\ $$
Answered by mindispower last updated on 02/Nov/21
t→^f 2 tg(t) is bijection,[0,(π/2)[→^f [0,+∞[ x=tg(t)⇔ ((cos(t))/(sin(t)))(((1−2sin(t)cos(t))/(cos^2 (x))))+sin(t)+cos(t)≥(√2) (1/(cos(t)sin(t)))+sin(t)+cos(t)≥2+(√2),∀t∈[0,(π/2)[ ((cos(t))/(sin(t)))+sin(t)+((sin(t))/(cos(t)))+cos(t)≥2+(√2) a=sin(t),b=cos(t) { ((a^2 +b^2 =1)),(((a/b)+(b/a)+a+b∣_(.=f(a,b)) ≥2+(√2))) :} ⇔(1/(ab))+a+b≥2+(√2) ⇔(1/(ab))+a+b≥2+2 f(a,b)=(1/(ab))+a+b−γ(a^2 +b^2 −1) (1/b)(−(1/a^2 ))+1−2γa=0 (1/a)(−(1/b^2 ))+1−2γb=0 (1/b)+(1/a)−2γ((a/b)+(b/a))=0 b γ=((a+b)/2) −(1/b^2 )+a−b+(1/a^2 )=0 b^2 −a^2 +a^2 b^2 (a−b)=0 −(a−b)(a+b+a^2 b^2 )=0 ⇒a=b=(1/( (√2))) Min(f)=f((1/( (√2))),(1/( (√2))))=(1/(1/2))+(1/( (√2)))+(1/( (√2)))=2+(√2)
$${t}\overset{{f}} {\rightarrow}\mathrm{2}\:{tg}\left({t}\right)\:{is}\:\:\:{bijection},\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\left[\overset{{f}} {\rightarrow}\left[\mathrm{0},+\infty\left[\right.\right.\right.\right. \\ $$$${x}={tg}\left({t}\right)\Leftrightarrow \\ $$$$\frac{{cos}\left({t}\right)}{{sin}\left({t}\right)}\left(\frac{\mathrm{1}−\mathrm{2}{sin}\left({t}\right){cos}\left({t}\right)}{{cos}^{\mathrm{2}} \left({x}\right)}\right)+{sin}\left({t}\right)+{cos}\left({t}\right)\geqslant\sqrt{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{cos}\left({t}\right){sin}\left({t}\right)}+{sin}\left({t}\right)+{cos}\left({t}\right)\geqslant\mathrm{2}+\sqrt{\mathrm{2}},\forall{t}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\left[\right.\right. \\ $$$$\frac{{cos}\left({t}\right)}{{sin}\left({t}\right)}+{sin}\left({t}\right)+\frac{{sin}\left({t}\right)}{{cos}\left({t}\right)}+{cos}\left({t}\right)\geqslant\mathrm{2}+\sqrt{\mathrm{2}} \\ $$$${a}={sin}\left({t}\right),{b}={cos}\left({t}\right) \\ $$$$\begin{cases}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1}}\\{\frac{{a}}{{b}}+\frac{{b}}{{a}}+{a}+{b}\underset{.={f}\left({a},{b}\right)} {\mid}\geqslant\mathrm{2}+\sqrt{\mathrm{2}}}\end{cases} \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{{ab}}+{a}+{b}\geqslant\mathrm{2}+\sqrt{\mathrm{2}} \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{{ab}}+{a}+{b}\geqslant\mathrm{2}+\mathrm{2} \\ $$$${f}\left({a},{b}\right)=\frac{\mathrm{1}}{{ab}}+{a}+{b}−\gamma\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{{b}}\left(−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)+\mathrm{1}−\mathrm{2}\gamma{a}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{a}}\left(−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)+\mathrm{1}−\mathrm{2}\gamma{b}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{a}}−\mathrm{2}\gamma\left(\frac{{a}}{{b}}+\frac{{b}}{{a}}\right)=\mathrm{0} \\ $$$${b} \\ $$$$ \\ $$$$\gamma=\frac{{a}+{b}}{\mathrm{2}} \\ $$$$−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+{a}−{b}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=\mathrm{0} \\ $$$${b}^{\mathrm{2}} −{a}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}−{b}\right)=\mathrm{0} \\ $$$$−\left({a}−{b}\right)\left({a}+{b}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{a}={b}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$${Min}\left({f}\right)={f}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\mathrm{2}+\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by HongKing last updated on 02/Nov/21
perfect my dear Ser, thank you
$$\mathrm{perfect}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{thank}\:\mathrm{you} \\ $$

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