Determine minimum value of ((sec^4 α)/(tan^2 β)) + ((sec^4 β)/(tan^2 α)) , over all α,β ≠ ((kπ)/2) and k∈Z


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Question Number 119372 by bobhans last updated on 24/Oct/20

$D e t e r m i n e m i n i m u m v a l u e o f$ $\frac{\sec^{4} α}{\tan^{2} β} + \frac{\sec^{4} β}{\tan^{2} α}, o v e r a l l α, β \neq \frac{k π}{2}$ $a n d k \in Z$
	Answered by 1549442205PVT last updated on 24/Oct/20
	$S= ((sec^4 α)/(tan^2 β)) + ((sec^4 β)/(tan^2 α)) = (1/(cos^4 α)).((cos^2 β)/(sin^2 β))+(1/(cos^4 β)).((cos^2 α)/(sin^2 α)) =(((1+tan^2 α)^2 )/(tan^2 β))+(((1+tan^2 β)^2 )/(tan^2 α)) Applying Cauchy−Shwardz we have S≥(((2+tan^2 α+tan^2 β)^2 )/(tan^2 α+tan^2 β)) =((4+(tan^2 α+tan^2 β)^2 +4(tan^2 α+tan^2 β))/(tan^2 α+tan^2 β)) =4+[tan^2 α+tan^2 β+(4/(tan^2 α+tan^2 β))] =4+((√(tan^2 α+tan^2 β))−(2/( (√(tan^2 α+tan^2 β)))))^2 +4≥8.The equality ocurrs if and only if { (((√(tan^2 α+tan^2 β))=(2/( (√(tan^2 α+tan^2 β))))(1))),((((1+tan^2 α)/(tan^2 β))=((1+tan^2 β)/(tan^2 α))(2))) :} (2)⇒tan^4 α+tan^2 α=tan^4 β+tan^2 β ⇔(tan^2 α−tan^2 β)(tan^2 α+tan^2 β+1)=0(∗) (1)⇒tan^2 α+tan^2 β=2.Replace into (∗) we get tan^2 α=tan^2 β=1 Thus,S= ((sec^4 α)/(tan^2 β)) + ((sec^4 β)/(tan^2 α)) has smallest value equal to 8 when tan^2 α=tan^2 β=1$
	$S = \frac{\sec^{4} α}{\tan^{2} β} + \frac{\sec^{4} β}{\tan^{2} α} = \frac{1}{\cos^{4} α} . \frac{\cos^{2} β}{\sin^{2} β} + \frac{1}{\cos^{4} β} . \frac{\cos^{2} α}{\sin^{2} α}$ $= \frac{{(1 + \tan^{2} α)}^{2}}{\tan^{2} β} + \frac{{(1 + \tan^{2} β)}^{2}}{\tan^{2} α}$ $Applying Cauchy - Shwardz we have$ $S ⩾ \frac{{(2 + \tan^{2} α + \tan^{2} β)}^{2}}{\tan^{2} α + \tan^{2} β}$ $= \frac{4 + {(\tan^{2} α + \tan^{2} β)}^{2} + 4 (\tan^{2} α + \tan^{2} β)}{\tan^{2} α + \tan^{2} β}$ $= 4 + [\tan^{2} α + \tan^{2} β + \frac{4}{\tan^{2} α + \tan^{2} β}]$ $= 4 + {(\sqrt{\tan^{2} α + \tan^{2} β} - \frac{2}{\sqrt{\tan^{2} α + \tan^{2} β}})}^{2}$ $+ 4 ⩾ 8 . The equality ocurrs if and only$ $if {\begin{cases} \sqrt{\tan^{2} α + \tan^{2} β} = \frac{2}{\sqrt{\tan^{2} α + \tan^{2} β}} (1) \\ \frac{1 + \tan^{2} α}{\tan^{2} β} = \frac{1 + \tan^{2} β}{\tan^{2} α} (2) \end{cases}$ $(2) \Rightarrow \tan^{4} α + \tan^{2} α = \tan^{4} β + \tan^{2} β$ $\Leftrightarrow (\tan^{2} α - \tan^{2} β) (\tan^{2} α + \tan^{2} β + 1) = 0 ()$ $(1) \Rightarrow \tan^{2} α + \tan^{2} β = 2 . Replace into$ $() we get \tan^{2} α = \tan^{2} β = 1$ $Thus, S = \frac{\sec^{4} α}{\tan^{2} β} + \frac{\sec^{4} β}{\tan^{2} α} has smallest$ $value equal to 8 when \tan^{2} α = \tan^{2} β = 1$
	Answered by bemath last updated on 24/Oct/20
	$set → { ((a=tan^2 α)),((b=tan^2 β)) :} it suffices to determine the minimum value of (((a+1)^2 )/b)+(((b+1)^2 )/a) with a,b≥0. we have (((a+1)^2 )/b)+(((b+1)^2 )/a)=((a^2 +2a+1)/b)+((b^2 +2b+1)/a) = ((a^2 /b)+(1/b)+(b^2 /a)+(1/a))+2((a/b)+(b/a)) ≥ 4 (((a^2 /b).(1/b).(b^2 /a).(1/a)))^(1/(4 )) + 4 (√((a/b).(b/a))) = 8$
	$s e t \to {\begin{cases} a = \tan^{2} α \\ b = \tan^{2} β \end{cases}$ $i t s u f f i c e s t o d e t e r m i n e t h e m i n i m u m v a l u e o f \frac{{(a + 1)}^{2}}{b} + \frac{{(b + 1)}^{2}}{a}$ $w i t h a, b ⩾ 0 . w e h a v e \frac{{(a + 1)}^{2}}{b} + \frac{{(b + 1)}^{2}}{a} = \frac{a^{2} + 2 a + 1}{b} + \frac{b^{2} + 2 b + 1}{a}$ $= (\frac{a^{2}}{b} + \frac{1}{b} + \frac{b^{2}}{a} + \frac{1}{a}) + 2 (\frac{a}{b} + \frac{b}{a})$ $⩾ 4 \sqrt[4]{\frac{a^{2}}{b} . \frac{1}{b} . \frac{b^{2}}{a} . \frac{1}{a}} + 4 \sqrt{\frac{a}{b} . \frac{b}{a}} = 8$
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