MHB Challenging Trigonometric Inequality: Can You Prove It?

[Albert1]

$0<\alpha, \beta<\dfrac {\pi}{2}$

prove :$\dfrac {1}{cos^2\alpha}+\dfrac {1}{sin^2\alpha\,sin^2\beta\, cos^2\beta}\geq 9$

and corresponding $\alpha$, and $\beta$

 

[kaliprasad]

$0<\alpha, \beta<\dfrac {\pi}{2}$

prove :$\dfrac {1}{cos^2\alpha}+\dfrac {1}{sin^2\alpha\,sin^2\beta\, cos^2\beta}\geq 9$

and corresponding $\alpha$, and $\beta$

Spoiler

$\dfrac{1}{\cos^2\alpha}+\dfrac{1}{\sin^2\alpha\sin^2\beta\cos^2\beta}$
= $\dfrac{1}{\cos^2\alpha}+\dfrac{4}{4\sin^2\alpha\sin^2\beta\cos^2\beta}$
= $\dfrac{1}{\cos^2\alpha}+\dfrac{1}{4\sin^2\alpha\sin^2 2 \beta}$
as $\alpha\, and,\beta$

are independedent we can take $\beta$ to be $\frac{\pi}{2}$

and we need to maximize $\sec^2 \alpha + 4 \csc^2 \alpha$
which is same as $\tan ^2 \alpha + 4 \cot^2 \alpha + 5 $
or $(\tan \alpha - 2 \cot\alpha)^2 + 9 $
which is $\ge 9 $

when $\alpha = \arctan \sqrt{2}$

 

[anemone]

$0<\alpha, \beta<\dfrac {\pi}{2}$

prove :$\dfrac {1}{cos^2\alpha}+\dfrac {1}{sin^2\alpha\,sin^2\beta\, cos^2\beta}\geq 9$

and corresponding $\alpha$, and $\beta$

My solution (kaliprasad beats me to it but I decided to go ahead and post my solution because we approached it differently and mine is less elegant, I think):

Spoiler

If we let $x=\sin^2 \alpha$, then $\cos^2 \alpha=1-x$ and similarly if we have $y=\sin^2 \beta$, then $\cos^2 \beta=1-y$

Note that since $0<\alpha, \beta<\dfrac {\pi}{2}$, this gives $0< x,\,y < 1$.

The given LHS of the inequality hence becomes

$\dfrac{1}{1-x}+\dfrac{1}{xy(1-y)}$

and bear in mind that we're now looking for its minimum value. We can rewrite it so that we have

$\begin{align*}\dfrac{1}{1-x}+\dfrac{1}{xy(1-y)}&=\dfrac{1}{1-x}-\dfrac{1}{xy(y-1)}\\&=\dfrac{1}{1-x}-\dfrac{1}{x((y-\dfrac{1}{2})^2-\dfrac{1}{4})}\\&\ge \dfrac{1}{1-x}+\dfrac{4}{x} \,\,\,\text{when $y=\dfrac{1}{2}$}\\&\ge\dfrac{3x-4}{x^2-x}\end{align*}$

The minimum of $\dfrac{3x-4}{x^2-x}$ can be found using the calculus method and this is obtained when $x=\dfrac{2}{3}$ where $\dfrac{3x-4}{x^2-x}=9$

Hence, $\dfrac {1}{\cos^2\alpha}+\dfrac {1}{\sin^2\alpha\,\sin^2\beta\, \cos^2\beta}\geq 9$ and equality is attained when $\sin^2 \alpha=\dfrac{2}{3}$ and $\sin^2 \beta=\dfrac{1}{2}$

 

[Albert1]

Spoiler

$\dfrac{1}{\cos^2\alpha}+\dfrac{1}{\sin^2\alpha\sin^2\beta\cos^2\beta}$
= $\dfrac{1}{\cos^2\alpha}+\dfrac{4}{4\sin^2\alpha\sin^2\beta\cos^2\beta}$
= $\dfrac{1}{\cos^2\alpha}+\dfrac{1}{4\sin^2\alpha\sin^2 2 \beta}---(1)$
as $\alpha\, and,\beta$
are independedent we can take $\beta$ to be $\frac{\pi}{2}$
and we need to maximize $\sec^2 \alpha + 4 \csc^2 \alpha$
which is same as $\tan ^2 \alpha + 4 \cot^2 \alpha + 5 $
or $(\tan \alpha - 2 \cot\alpha)^2 + 9 $
which is $\ge 9 $

when $\alpha = \arctan \sqrt{2}$

$\dfrac{1}{\cos^2\alpha}+\dfrac{1}{4\sin^2\alpha\sin^2 2 \beta}---(1)$
you take $\beta$ to be $\frac{\pi}{2}$
then from (1) $sin 2\beta =0$
the denominator of the second term in (1) will be 0

I think you should take $\beta$ to be $\frac{\pi}{4}$

 

[kaliprasad]

$\dfrac{1}{\cos^2\alpha}+\dfrac{1}{4\sin^2\alpha\sin^2 2 \beta}---(1)$
you take $\beta$ to be $\frac{\pi}{2}$
then from (1) $sin 2\beta =0$
the denominator of the second term in (1) will be 0

I think you should take $\beta$ to be $\frac{\pi}{4}$

yes you are right. it was my inadvertent error. Thanks for correcting

 

[kaliprasad]

opalg has beaten me by months
http://mathhelpboards.com/challenge-questions-puzzles-28/trigonometric-inequality-9277.html