∀ a,b>0 , a^2 +b^2 =1 prove that ((1/a)+(1/b))((b/(a^2 +1))+(a/(b^2 +1)))≥(8/3)


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Question Number 98267 by M±th+et+s last updated on 12/Jun/20

$\forall a, b > 0, a^{2} + b^{2} = 1$ $p r o v e t h a t$ $(\frac{1}{a} + \frac{1}{b}) (\frac{b}{a^{2} + 1} + \frac{a}{b^{2} + 1}) ⩾ \frac{8}{3}$
	Answered by maths mind last updated on 13/Jun/20

	$(\frac{a + b}{a b}) (\frac{b^{3} + b + a^{3} + a}{2 + a^{2} b^{2}})$ $(b^{3} + a^{3} + a + b) = (a + b) (b^{2} + a^{2} - a b + 1)$ $= (a + b) (2 - a b)$ $\Leftrightarrow \frac{{(a + b)}^{2} (2 - a b)}{a b (2 + a^{2} b^{2})}$ ${(a + b)}^{2} = 1 + 2 a b$ $\Leftrightarrow \frac{(1 + 2 a b) (2 - a b)}{a b (2 + a^{2} b^{2})} ⩾ \frac{8}{3}$ $l e t a b = x \Leftrightarrow \frac{(1 + 2 x) (2 - x)}{x (2 + x^{2})} ⩾ \frac{8}{3}$ $\Leftrightarrow 8 x^{3} + 6 x^{2} + 7 x - 6 ⩽ 0$ $a b ⩽ \frac{1}{2} (a^{2} + b^{2}) = \frac{1}{2} \Rightarrow$ $8 x^{3} ⩽ 1, 7 x ⩽ \frac{7}{2}, 6 x^{2} ⩽ \frac{3}{2}$ $\Rightarrow 8 x^{3} + 6 x^{2} + 7 x - 6 ⩽ 1 + \frac{7}{2} + \frac{3}{2} - 6 = 0$ $\Leftrightarrow (\frac{1}{a} + \frac{1}{b}) (\frac{a}{b^{2} + 1} + \frac{b}{a^{2} + 1}) ⩾ \frac{8}{3}$
	Commented by M±th+et+s last updated on 13/Jun/20
	�� great work
	Commented bymaths mind last updated on 13/Jun/20

	$t h a n x, i g o t a n s w e r f o r y o u r p o s t$ $i w i l l p o s t i t l a t t e r$
	Commented by M±th+et+s last updated on 13/Jun/20

	$t h a n k y o u s o m u c h .$ $g o d b l e s s y o u s i r$
	Answered by 1549442205 last updated on 13/Jun/20
	$Putting a=cosϕ,b=sinϕ,ϕ∈[0;(π/2)].Then the given inequality becomes ((1/(cosϕ))+(1/(sinϕ)))(((sinϕ)/(cos^2 ϕ+1))+((cosϕ)/(sin^2 ϕ+1)))≥(8/3) ⇔(((cosϕ+sinϕ)(cos^3 ϕ+sin^3 ϕ+cosϕ+sinϕ))/(cosϕ.sinϕ(2+cos^2 ϕsin^2 ϕ)))≥(8/3) ⇔(((cosϕ+sinϕ)^2 (2−sinϕcosϕ))/(((sin2ϕ)/2).(2+((sin^2 ϕ)/4))))≥(8/3) ⇔(((1+sin2ϕ)(2−((sin2ϕ)/2)))/(sin2ϕ+((sin^3 ϕ)/8)))≥(8/3) ⇔3{2+(3/2)sin2ϕ−((sin^2 2ϕ)/2)}≥8sin2ϕ+sin^3 2ϕ ⇔sin^3 2ϕ+(3/2)sin^2 2ϕ+(7/2)sin2ϕ−6≤0 ⇔2sin^3 2ϕ+3sin^2 2ϕ+7sin2ϕ−12≤0 ⇔(sin2ϕ−1)(2sin^2 2ϕ+5sin2ϕ+12)≤0 The final inequality is always true because (2sin^2 2ϕ+5sin2ϕ+12)=2(sin2ϕ+(5/4))^2 +((71)/8)>0 and sin2ϕ≤1.Therefore the given inequality proved.The equality occurs if and only if sin2ϕ=1⇔a=b=((√2)/2)$
	$Putting a = \cos φ, b = \sin φ, φ \in [0; \frac{π}{2}] . Then$ $the given inequality becomes$ $(\frac{1}{\cos φ} + \frac{1}{\sin φ}) (\frac{\sin φ}{\cos^{2} φ + 1} + \frac{\cos φ}{\sin^{2} φ + 1}) ⩾ \frac{8}{3}$ $\Leftrightarrow \frac{(\cos φ + \sin φ) (\cos^{3} φ + \sin^{3} φ + \cos φ + \sin φ)}{\cos φ . \sin φ (2 + \cos^{2} φ \sin^{2} φ)} ⩾ \frac{8}{3}$ $\Leftrightarrow \frac{{(\cos φ + \sin φ)}^{2} (2 - \sin φ \cos φ)}{\frac{\sin 2 φ}{2} . (2 + \frac{\sin^{2} φ}{4})} ⩾ \frac{8}{3}$ $\Leftrightarrow \frac{(1 + \sin 2 φ) (2 - \frac{\sin 2 φ}{2})}{\sin 2 φ + \frac{\sin^{3} φ}{8}} ⩾ \frac{8}{3}$ $\Leftrightarrow 3 {2 + \frac{3}{2} \sin 2 φ - \frac{\sin^{2} 2 φ}{2}} ⩾ 8 \sin 2 φ + \sin^{3} 2 φ$ $\Leftrightarrow \sin^{3} 2 φ + \frac{3}{2} \sin^{2} 2 φ + \frac{7}{2} \sin 2 φ - 6 ⩽ 0$ $\Leftrightarrow {2 \sin}^{3} 2 φ + {3 \sin}^{2} 2 φ + 7 \sin 2 φ - 12 ⩽ 0$ $\Leftrightarrow (\sin 2 φ - 1) ({2 \sin}^{2} 2 φ + 5 \sin 2 φ + 12) ⩽ 0$ $The final inequality is always true because$ $({2 \sin}^{2} 2 φ + 5 \sin 2 φ + 12) = 2 {(\sin 2 φ + \frac{5}{4})}^{2} + \frac{71}{8} > 0$ $and \sin 2 φ ⩽ 1 . Therefore the given$ $inequality proved . The equality occurs$ $if and only if \sin 2 φ = 1 \Leftrightarrow a = b = \frac{\sqrt{2}}{2}$
	Commented by M±th+et+s last updated on 13/Jun/20

	$t h a n k y o u s i r$
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