Question Number 119372 by bobhans last updated on 24/Oct/20
$${Determine}\:{minimum}\:{value}\:{of}\: \\ $$$$\:\frac{\mathrm{sec}\:^{\mathrm{4}} \alpha}{\mathrm{tan}\:^{\mathrm{2}} \beta}\:+\:\frac{\mathrm{sec}\:^{\mathrm{4}} \beta}{\mathrm{tan}\:^{\mathrm{2}} \alpha}\:,\:{over}\:{all}\:\alpha,\beta\:\neq\:\frac{{k}\pi}{\mathrm{2}} \\ $$$${and}\:{k}\in\mathbb{Z} \\ $$
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](https://www.tinkutara.com/question/Q119419.png)
$$\mathrm{S}=\:\frac{\mathrm{sec}\:^{\mathrm{4}} \alpha}{\mathrm{tan}\:^{\mathrm{2}} \beta}\:+\:\frac{\mathrm{sec}\:^{\mathrm{4}} \beta}{\mathrm{tan}\:^{\mathrm{2}} \alpha}\:=\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{4}} \alpha}.\frac{\mathrm{cos}^{\mathrm{2}} \beta}{\mathrm{sin}^{\mathrm{2}} \beta}+\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{4}} \beta}.\frac{\mathrm{cos}^{\mathrm{2}} \alpha}{\mathrm{sin}^{\mathrm{2}} \alpha} \\ $$$$=\frac{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \alpha\right)^{\mathrm{2}} }{\mathrm{tan}^{\mathrm{2}} \beta}+\frac{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \beta\right)^{\mathrm{2}} }{\mathrm{tan}^{\mathrm{2}} \alpha} \\ $$$$\mathrm{Applying}\:\mathrm{Cauchy}−\mathrm{Shwardz}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{S}\geqslant\frac{\left(\mathrm{2}+\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{tan}^{\mathrm{2}} \beta\right)^{\mathrm{2}} }{\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{tan}^{\mathrm{2}} \beta} \\ $$$$=\frac{\mathrm{4}+\left(\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{tan}^{\mathrm{2}} \beta\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{tan}^{\mathrm{2}} \beta\right)}{\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{tan}^{\mathrm{2}} \beta} \\ $$$$=\mathrm{4}+\left[\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{tan}^{\mathrm{2}} \beta+\frac{\mathrm{4}}{\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{tan}^{\mathrm{2}} \beta}\right] \\ $$$$=\mathrm{4}+\left(\sqrt{\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{tan}^{\mathrm{2}} \beta}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{tan}^{\mathrm{2}} \beta}}\right)^{\mathrm{2}} \\ $$$$+\mathrm{4}\geqslant\mathrm{8}.\mathrm{The}\:\mathrm{equality}\:\mathrm{ocurrs}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only} \\ $$$$\mathrm{if}\:\begin{cases}{\sqrt{\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{tan}^{\mathrm{2}} \beta}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{tan}^{\mathrm{2}} \beta}}\left(\mathrm{1}\right)}\\{\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \alpha}{\mathrm{tan}^{\mathrm{2}} \beta}=\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \beta}{\mathrm{tan}^{\mathrm{2}} \alpha}\left(\mathrm{2}\right)}\end{cases} \\ $$$$\left(\mathrm{2}\right)\Rightarrow\mathrm{tan}^{\mathrm{4}} \alpha+\mathrm{tan}^{\mathrm{2}} \alpha=\mathrm{tan}^{\mathrm{4}} \beta+\mathrm{tan}^{\mathrm{2}} \beta \\ $$$$\Leftrightarrow\left(\mathrm{tan}^{\mathrm{2}} \alpha−\mathrm{tan}^{\mathrm{2}} \beta\right)\left(\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{tan}^{\mathrm{2}} \beta+\mathrm{1}\right)=\mathrm{0}\left(\ast\right) \\ $$$$\left(\mathrm{1}\right)\Rightarrow\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{tan}^{\mathrm{2}} \beta=\mathrm{2}.\mathrm{Replace}\:\mathrm{into} \\ $$$$\left(\ast\right)\:\mathrm{we}\:\mathrm{get}\:\mathrm{tan}^{\mathrm{2}} \alpha=\mathrm{tan}^{\mathrm{2}} \beta=\mathrm{1} \\ $$$$\mathrm{Thus},\mathrm{S}=\:\frac{\mathrm{sec}\:^{\mathrm{4}} \alpha}{\mathrm{tan}\:^{\mathrm{2}} \beta}\:+\:\frac{\mathrm{sec}\:^{\mathrm{4}} \beta}{\mathrm{tan}\:^{\mathrm{2}} \alpha}\:\:\mathrm{has}\:\mathrm{smallest} \\ $$$$\mathrm{value}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{8}\:\mathrm{when}\:\mathrm{tan}^{\mathrm{2}} \alpha=\mathrm{tan}^{\mathrm{2}} \beta=\mathrm{1} \\ $$
Answered by bemath last updated on 24/Oct/20
$${set}\:\rightarrow\begin{cases}{{a}=\mathrm{tan}\:^{\mathrm{2}} \alpha}\\{{b}=\mathrm{tan}\:^{\mathrm{2}} \beta}\end{cases} \\ $$$${it}\:{suffices}\:{to}\:{determine}\:{the}\:{minimum}\:{value}\:{of}\:\:\frac{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }{{b}}+\frac{\left({b}+\mathrm{1}\right)^{\mathrm{2}} }{{a}} \\ $$$${with}\:{a},{b}\geqslant\mathrm{0}.\:{we}\:{have}\:\frac{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }{{b}}+\frac{\left({b}+\mathrm{1}\right)^{\mathrm{2}} }{{a}}=\frac{{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}}{{b}}+\frac{{b}^{\mathrm{2}} +\mathrm{2}{b}+\mathrm{1}}{{a}} \\ $$$$\:=\:\left(\frac{{a}^{\mathrm{2}} }{{b}}+\frac{\mathrm{1}}{{b}}+\frac{{b}^{\mathrm{2}} }{{a}}+\frac{\mathrm{1}}{{a}}\right)+\mathrm{2}\left(\frac{{a}}{{b}}+\frac{{b}}{{a}}\right) \\ $$$$\:\geqslant\:\mathrm{4}\:\sqrt[{\mathrm{4}\:}]{\frac{{a}^{\mathrm{2}} }{{b}}.\frac{\mathrm{1}}{{b}}.\frac{{b}^{\mathrm{2}} }{{a}}.\frac{\mathrm{1}}{{a}}}\:+\:\mathrm{4}\:\sqrt{\frac{{a}}{{b}}.\frac{{b}}{{a}}}\:=\:\mathrm{8} \\ $$