Prove This beautiful result !

$\alpha \quad ,\quad \beta \quad >\quad 0\quad \quad \& \quad \quad \theta \quad \in \quad R\\ \\ f\left( \theta \right) \quad =\quad \quad \alpha \sec ^{ 2 }{ \theta } \quad +\quad \beta \csc ^{ 2 }{ \theta } \quad +\quad \sqrt { { \alpha }^{ 2 }\sec ^{ 4 }{ \theta } \quad +\quad { \beta }^{ 2 }\csc ^{ 4 }{ \theta } }$ .

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

${ f\left( \theta \right) }_{ min }\quad =\quad 2(\alpha \quad +\quad \beta \quad +\quad \sqrt { 2\alpha \beta } )$ .

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, !!

#Geometry #Minimum #GeometricInterpretation #Beauty-of-Maths

Easy Math Editor

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Okay Now its time to give Hint :

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

Deepanshu Gupta - 4 years, 11 months ago

Ohh ! Now I got something

is it mean :

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

I think it is Perimeter of right angle triangle ?

Am i going Correct ?

Karan Shekhawat - 4 years, 11 months ago

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

Deepanshu Gupta - 4 years, 11 months ago

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

Wijil Pambudi - 4 years, 11 months ago

See Unravelling an inequality problem by Calvin Lin.

Souryajit Roy - 4 years, 11 months ago

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

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

imgur imgur

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

also Let $\sqrt { { a }^{ 2 }\quad +\quad { b }^{ 2 } } \quad =\quad { C }_{ 1 }\quad +\quad { C }_{ 2 }$ .

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

Now For Minimum perimeter P we use AM-GM inequality

$P\quad \ge \quad 2\sqrt { (\quad a\quad +\quad { C }_{ 2 }\quad )(\quad b\quad +\quad { C }_{ 1 }\quad ) } \\ \\ { P }_{ min }\quad =\quad 2\sqrt { (\quad a\quad +\quad { C }_{ 2 }\quad )(\quad b\quad +\quad { C }_{ 1 }\quad ) } \quad \\$

If and only iff :

1)- GM must be constant

2)- Equality of variable must be attained

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

Now Same as Yours !

Is This is correct ? Calvin Sir ?

Deepanshu Gupta - 4 years, 11 months ago

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

sandeep Rathod - 4 years, 11 months ago

@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

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

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

Karan Shekhawat - 4 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$