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

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, !!

## Comments

Sort by:

TopNewestgood Question,,!! rememberedme to high school..!! :D – Wijil Pambudi · 1 year, 10 months ago

Log in to reply

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 · 1 year, 10 months ago

Log in to reply

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 · 1 year, 10 months ago

Log in to reply

– Deepanshu Gupta · 1 year, 10 months ago

Yes ! You are on right Path ! Keep Trying You will find the way :)Log in to reply

See Unravelling an inequality problem by Calvin Lin. – Souryajit Roy · 1 year, 10 months ago

Log in to reply

Recognizing and relating patterns is helpful in approaching such problems. – Calvin Lin Staff · 1 year, 10 months ago

Log in to reply

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

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 · 1 year, 10 months ago

Log in to reply

@Deepanshu Gupta – Sandeep Rathod · 1 year, 10 months ago

what about area at this stage?Log in to reply

@sandeep Rathod – Deepanshu Gupta · 1 year, 10 months ago

I don't think that we can do more for calculating area from above information in this particular question . what do you think ?Log in to reply

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

But i solved it in

slightlydifferent manner as that calvin sir posted ( also my friend also help me little in how to starting this problem ) – Deepanshu Gupta · 1 year, 10 months agoLog in to reply

@Deepanshu Gupta – Karan Shekhawat · 1 year, 10 months ago

So what is your Slightly different approach Can you post it ! please Thanks !Log in to reply

@KARAN SHEKHAWAT

I had Posted itBut I'am not 100% sure that it is correct or not ! – Deepanshu Gupta · 1 year, 10 months ago

Log in to reply

Log in to reply

@Sandeep Bhardwaj @Sanjeet Raria – Megh Choksi · 1 year, 10 months ago

Log in to reply

– Megh Choksi · 1 year, 10 months ago

I am too trying it with \( \sqrt{\frac{x_1^2+\cdots+x_n^2}{n}}\ge\frac{x_1+\cdots+x_n}{n}\) for the third term.Log in to reply