Prove that: (((x-1)^2 )/x) + ((x+1)/( (√(x^2 +1)))) ≥ (√2) ; ∀x>0


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Question Number 158245 by HongKing last updated on 01/Nov/21

$Prove that :$ $\frac{{(x - 1)}^{2}}{x} + \frac{x + 1}{\sqrt{x^{2} + 1}} ⩾ \sqrt{2}; \forall x > 0$
	Answered by mindispower last updated on 02/Nov/21
	$t→^f 2 tg(t) is bijection,[0,(π/2)[→^f [0,+∞[ x=tg(t)⇔ ((cos(t))/(sin(t)))(((1−2sin(t)cos(t))/(cos^2 (x))))+sin(t)+cos(t)≥(√2) (1/(cos(t)sin(t)))+sin(t)+cos(t)≥2+(√2),∀t∈[0,(π/2)[ ((cos(t))/(sin(t)))+sin(t)+((sin(t))/(cos(t)))+cos(t)≥2+(√2) a=sin(t),b=cos(t) { ((a^2 +b^2 =1)),(((a/b)+(b/a)+a+b∣_(.=f(a,b)) ≥2+(√2))) :} ⇔(1/(ab))+a+b≥2+(√2) ⇔(1/(ab))+a+b≥2+2 f(a,b)=(1/(ab))+a+b−γ(a^2 +b^2 −1) (1/b)(−(1/a^2 ))+1−2γa=0 (1/a)(−(1/b^2 ))+1−2γb=0 (1/b)+(1/a)−2γ((a/b)+(b/a))=0 b γ=((a+b)/2) −(1/b^2 )+a−b+(1/a^2 )=0 b^2 −a^2 +a^2 b^2 (a−b)=0 −(a−b)(a+b+a^2 b^2 )=0 ⇒a=b=(1/( (√2))) Min(f)=f((1/( (√2))),(1/( (√2))))=(1/(1/2))+(1/( (√2)))+(1/( (√2)))=2+(√2)$
	$t \overset{f}{\to} 2 t g (t) i s b i j e c t i o n, [0, \frac{π}{2} [\overset{f}{\to} [0, + \infty [$ $x = t g (t) \Leftrightarrow$ $\frac{c o s (t)}{s i n (t)} (\frac{1 - 2 s i n (t) c o s (t)}{{c o s}^{2} (x)}) + s i n (t) + c o s (t) ⩾ \sqrt{2}$ $\frac{1}{c o s (t) s i n (t)} + s i n (t) + c o s (t) ⩾ 2 + \sqrt{2}, \forall t \in [0, \frac{π}{2} [$ $\frac{c o s (t)}{s i n (t)} + s i n (t) + \frac{s i n (t)}{c o s (t)} + c o s (t) ⩾ 2 + \sqrt{2}$ $a = s i n (t), b = c o s (t)$ ${\begin{cases} a^{2} + b^{2} = 1 \\ \frac{a}{b} + \frac{b}{a} + a + b \underset{. = f (a, b)}{∣} ⩾ 2 + \sqrt{2} \end{cases}$ $\Leftrightarrow \frac{1}{a b} + a + b ⩾ 2 + \sqrt{2}$ $\Leftrightarrow \frac{1}{a b} + a + b ⩾ 2 + 2$ $f (a, b) = \frac{1}{a b} + a + b - γ (a^{2} + b^{2} - 1)$ $\frac{1}{b} (- \frac{1}{a^{2}}) + 1 - 2 γ a = 0$ $\frac{1}{a} (- \frac{1}{b^{2}}) + 1 - 2 γ b = 0$ $\frac{1}{b} + \frac{1}{a} - 2 γ (\frac{a}{b} + \frac{b}{a}) = 0$ $b$ $γ = \frac{a + b}{2}$ $- \frac{1}{b^{2}} + a - b + \frac{1}{a^{2}} = 0$ $b^{2} - a^{2} + a^{2} b^{2} (a - b) = 0$ $- (a - b) (a + b + a^{2} b^{2}) = 0$ $\Rightarrow a = b = \frac{1}{\sqrt{2}}$ $M i n (f) = f (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) = \frac{1}{\frac{1}{2}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = 2 + \sqrt{2}$
	Commented by HongKing last updated on 02/Nov/21

	$perfect my dear Ser, thank you$
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