MHB Prove The Product Is Greater Than 5

[anemone]

Prove $$\left(1+\frac{1}{\sin x}\right)\left(1+\frac{1}{\cos x}\right)\gt 5$$ for $$0\lt x \lt \frac{\pi}{2}$$.

 

[Euge]

Hi anemone,

Here is my solution.

Spoiler

I'll prove the stronger statement

$$\left(1 + \frac{1}{\sin x}\right)\left(1 + \frac{1}{\cos x}\right) \ge 3 + 2\sqrt{2} \qquad (0 < x < \frac{\pi}{2})$$

This inequality is stronger than the proposed one since $3 + 2\sqrt{2} > 3 +2 = 5$. Expand the product on the left-hand side of the inequality to get

$$1 + \frac{1}{\sin x} + \frac{1}{\cos x} + \frac{1}{\sin x\cos x}\tag{*}$$By the arithmetic-harmonic mean inequality,

$$\frac{1}{\sin x} + \frac{1}{\cos x} \ge \frac{4}{\sin x + \cos x} = \frac{4}{\sqrt{2}\sin(x + \pi/4)} \le \frac{4}{\sqrt{2}} = 2\sqrt{2}$$

Since

$$\frac{1}{\sin x\cos x} = \frac{2}{\sin 2x} \ge 2$$

we deduce that the expression (*) is at least $1 + 2\sqrt{2} + 2$, or $3 + 2\sqrt{2}$. Note that equality holds if and only if $x = \pi/4$.

 

[MarkFL]

Prove $$\left(1+\frac{1}{\sin x}\right)\left(1+\frac{1}{\cos x}\right)\gt 5$$ for $$0\lt x \lt \frac{\pi}{2}$$.

My solution:

Spoiler

Let the objective function be:

$$f(x,y)=(1+\csc(x))(1+\csc(y))$$

Subject to the constraint:

$$g(x,y)=x+y-\frac{\pi}{2}=0$$ where $$0<x,y<\frac{\pi}{2}$$

Now, by cyclic symmetry, we find the critical point is at:

$$(x,y)=\left(\frac{\pi}{4},\frac{\pi}{4}\right)$$

And we also find:

$$f\left(\frac{\pi}{4},\frac{\pi}{4}\right)=\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}$$

Now, if we pick another point on the constraint, such as:

$$(x,y)=\left(\frac{\pi}{6},\frac{\pi}{3}\right)$$

We then find

$$f\left(\frac{\pi}{6},\frac{\pi}{3}\right)=(1+2)\left(1+\frac{2}{\sqrt{3}}\right)=\sqrt{3}\left(2+\sqrt{3}\right)=3+2\sqrt{3}>3+2\sqrt{2}$$

And so we conclude that:

$$f_{\min}=3+2\sqrt{2}>5$$

 

[Albert1]

Prove $$\left(1+\frac{1}{\sin x}\right)\left(1+\frac{1}{\cos x}\right)\gt 5---(1)$$ for $$0\lt x \lt \frac{\pi}{2}$$.

my solution:

Spoiler

using $AP\geq GP$
$(1)> 4{\sqrt {\dfrac{2}{(sin\,2x)}}}>4\sqrt 2>5$

 

[anemone]

Hi Euge, MarkFL and Albert!

Very good job to the three of you! And thanks for participating!