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Prove the inequality: $$\left(1+\dfrac{1}{\sin a}\right)\left(1+\dfrac{1}{\cos a}\right)\ge 3+2\sqrt{2}; \text{ for } a\in\left]0,\frac{\pi}{2}\right[$$

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  • $\begingroup$ what is $a$ here? $\endgroup$ Dec 27, 2015 at 19:44
  • $\begingroup$ Simply not true. When $\sin a=\cos a=-\frac{\sqrt{2}}{2}$, for example. $\endgroup$ Dec 27, 2015 at 19:44
  • $\begingroup$ is it $(1+\frac{1}{\sin(a)})$ or $\frac{1}{1+\sin(a)}$? $\endgroup$ Dec 27, 2015 at 19:45
  • $\begingroup$ a should be from 0 to pi $\endgroup$ Dec 27, 2015 at 19:45
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    $\begingroup$ Probably $[0,\pi]$ isn't true, either, since $\sin a=\frac{\sqrt 2}{2},\cos a=-\frac{\sqrt{2}}{2}$ is going to give a negative value for the product. You probably want $a\in(0,\pi/2)$. But just guessing. $\endgroup$ Dec 27, 2015 at 19:50

4 Answers 4

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By the AM-GM inequality, $$\dfrac{1}{\cos a}+\dfrac{1}{\sin a} \geq 2 \sqrt{\dfrac{1}{\sin a\cos a}}$$

Since $\sin a \cos a = \frac12 \sin(2a) \leq \frac12$, we have $\dfrac{1}{\sin a\cos a}\geq 2$.

Hence $$\left(1+\dfrac{1}{\sin a}\right)\left(1+\dfrac{1}{\cos a}\right) = 1+\dfrac{1}{\cos a}+\dfrac{1}{\sin a}+\dfrac{1}{\sin a\cos a} \\\geq 1+2\sqrt{\dfrac{1}{\sin a\cos a}}+\dfrac{1}{\sin a\cos a}\geq 1+2\sqrt{2} + 2 = 3+ 2\sqrt{2}$$

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Here is another proof:

Assume $x^2+y^2=1$ then $(x+y)^2=1+2xy$ and $$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\frac{xy+x+y+1}{xy}=\frac{(x+y+1)^2}{(x+y)^2-1}=\frac{x+y+1}{x+y-1}$$ Now rearranging the inequality becomes $x+y\leq \sqrt{2}$. And it can be seen simply that this is the maximum value of $x+y$ given $x^2+y^2=1$, for example $z=x+y$ is a plane and its intersection with the cylinder $x^2+y^2=1$has a maximum at $x=y$.

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I think it should be $(0,\frac{\pi}{2})$, so that both $\sin a$ and $\cos a$ are positive and well defined.

$$\bigg (1+\frac{1}{\sin a} \bigg)\bigg (1+\frac{1}{\cos a} \bigg) = 1+ \frac{1}{\sin a} + \frac{1}{\cos a} + \frac{1}{\sin a\cos a}$$

$$\frac{1}{\sin a} + \frac{1}{\cos a} \geq \frac{4}{\sin a + \cos a} \geq \frac{4}{\sqrt{2(\sin^2 a + \cos^2 a)}} = 2\sqrt 2$$

$$\frac{1}{\sin a\cos a} \geq \frac{2}{\sin^2 a + \cos^2 a} = 2$$

So, $$\bigg (1+\frac{1}{\sin a} \bigg)\bigg (1+\frac{1}{\cos a} \bigg) = 1+ \frac{1}{\sin a} + \frac{1}{\cos a} + \frac{1}{\sin a\cos a} \geq 3+ 2\sqrt 2$$

Equality occurs when $a = \frac{\pi}{4}$

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Differentiate to find the minimum of the LHS :

$$\frac{\mathrm{d}}{\mathrm{d}a}\left(\,\left(1+\frac{1}{\sin\,a}\right)\left(1+\frac{1}{\cos\,a}\right)\,\right)=0$$

$$\left(-\frac{1}{\sin^2 a}\right)\cos a\left(1+\frac{1}{\cos a}\right)+\left(1+\frac{1}{\sin a}\right)(-\sin a)\left(-\frac{1}{\cos^2 a}\right)=0$$

$$\left(\sin\,a+1\right)\left(-\frac{1}{\cos^2 a}\right)=\left(\cos\,a+1\right)\left(\frac{1}{\sin^2 a}\right)$$

$$(1+\sin a)(\sin^2a)=(1+\cos a)(\cos^2a)=(1+\cos a)(1-\sin^2a)$$

$$(1+\sin\,a)(\sin^2a)=(1+\cos\,a)(1+\sin\,a)(1-\sin\,a)$$

$$\sin^2a=1-\cos^2a=(1-\cos\,a)(1+\cos\,a)=(1+\cos\,a)(1-\sin\,a)$$

$$1-\cos\,a=1-\sin\,a$$

$$\cos\,a=\sin\,a$$

This only happens at $a=\frac{\pi}{4}$ in the given range.

And here $\sin\,a=\cos\,a=\frac{1}{\sqrt{2}}$.

So the minimum of the LHS is $(1+\sqrt{2})^2=1+2+2\sqrt{2}=3+2\sqrt{2}$

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