Prove This beautiful result !

α,β>0&θRf(θ)=αsec2θ+βcsc2θ+α2sec4θ+β2csc4θ\alpha \quad ,\quad \beta \quad >\quad 0\quad \quad \& \quad \quad \theta \quad \in \quad R\\ \\ f\left( \theta \right) \quad =\quad \quad \alpha \sec ^{ 2 }{ \theta } \quad +\quad \beta \csc ^{ 2 }{ \theta } \quad +\quad \sqrt { { \alpha }^{ 2 }\sec ^{ 4 }{ \theta } \quad +\quad { \beta }^{ 2 }\csc ^{ 4 }{ \theta } } .

Then Prove That Minimum Value of f(θ)f\left( \theta \right) is :

f(θ)min=2(α+β+2αβ){ f\left( \theta \right) }_{ min }\quad =\quad 2(\alpha \quad +\quad \beta \quad +\quad \sqrt { 2\alpha \beta } ).


Extra credit: Just observe it !
( It's Look ugly But Believe me in actual it is very very Beautiful question )

Following tool of mathematics are may Helpful Here Like : Substitution , Geometrical interpretation , AM - GM , Circles , Straight lines , Polar substitution etc, !!

Note by Deepanshu Gupta
4 years, 11 months ago

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Okay Now its time to give Hint :

use Substitution : a=αsec2θb=βcsc2θa\quad =\quad \alpha \sec ^{ 2 }{ \theta } \\ \\ b\quad =\quad \beta \csc ^{ 2 }{ \theta } .

Deepanshu Gupta - 4 years, 11 months ago

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Ohh ! Now I got something

is it mean :

f(θ)=a+b+a2+b2f(\theta )\quad =\quad a\quad +\quad b\quad +\quad \sqrt { { a }^{ 2 }\quad +\quad { b }^{ 2 } } .

I think it is Perimeter of right angle triangle ?

Am i going Correct ?

Karan Shekhawat - 4 years, 11 months ago

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Yes ! You are on right Path ! Keep Trying You will find the way :)

Deepanshu Gupta - 4 years, 11 months ago

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good Question,,!! rememberedme to high school..!! :D

Wijil Pambudi - 4 years, 11 months ago

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See Unravelling an inequality problem by Calvin Lin.

Souryajit Roy - 4 years, 11 months ago

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That's certainly one way to motivate the solution.

Recognizing and relating patterns is helpful in approaching such problems.

Calvin Lin Staff - 4 years, 11 months ago

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@Calvin Lin Sir I solved in this way :(Similar to your Awesome Technique )

imgur imgur

Let P=a+b+a2+b2P\quad =\quad a\quad +\quad b\quad +\quad \sqrt { { a }^{ 2 }\quad +\quad { b }^{ 2 } } .

also Let a2+b2=C1+C2\sqrt { { a }^{ 2 }\quad +\quad { b }^{ 2 } } \quad =\quad { C }_{ 1 }\quad +\quad { C }_{ 2 }.

P=(a+C2)+(b+C1)P\quad =\quad (\quad a\quad +\quad { C }_{ 2 }\quad )\quad +\quad (\quad b\quad +\quad { C }_{ 1 }\quad )\quad .

Now For Minimum perimeter P we use AM-GM inequality

P2(a+C2)(b+C1)Pmin=2(a+C2)(b+C1)P\quad \ge \quad 2\sqrt { (\quad a\quad +\quad { C }_{ 2 }\quad )(\quad b\quad +\quad { C }_{ 1 }\quad ) } \\ \\ { P }_{ min }\quad =\quad 2\sqrt { (\quad a\quad +\quad { C }_{ 2 }\quad )(\quad b\quad +\quad { C }_{ 1 }\quad ) } \quad \\

If and only iff :

1)- GM must be constant

2)- Equality of variable must be attained

(a+C2)(b+C1)=constant(a+C2)=(b+C1)=r(say)Pmin=2r\\ \bullet \quad (\quad a\quad +\quad { C }_{ 2 }\quad )(\quad b\quad +\quad { C }_{ 1 }\quad )\quad =\quad constant\\ \\ \bullet \quad (\quad a\quad +\quad { C }_{ 2 }\quad )\quad =\quad (\quad b\quad +\quad { C }_{ 1 }\quad )\quad =\quad r\quad (say)\\ \\ { P }_{ min }\quad =\quad 2r\quad .

Now Same as Yours !

Is This is correct ? Calvin Sir ?

Deepanshu Gupta - 4 years, 11 months ago

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@Deepanshu Gupta what about area at this stage?@Deepanshu Gupta

sandeep Rathod - 4 years, 11 months ago

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@Sandeep Rathod I don't think that we can do more for calculating area from above information in this particular question . what do you think ? @sandeep Rathod

Deepanshu Gupta - 4 years, 11 months ago

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ohh is it a famous problem? I don't know about this . Actually this question was giving to me by a friend of me as a challange !

So I Liked it after solving it So i share this ! ,

But i solved it in slightly different manner as that calvin sir posted ( also my friend also help me little in how to starting this problem )

Deepanshu Gupta - 4 years, 11 months ago

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So what is your Slightly different approach Can you post it ! please Thanks ! @Deepanshu Gupta

Karan Shekhawat - 4 years, 11 months ago

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@Karan Shekhawat I had Posted it @KARAN SHEKHAWAT

But I'am not 100% sure that it is correct or not !

Deepanshu Gupta - 4 years, 11 months ago

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