How to prove $\left(1+\frac{1}{\sin a}\right)\left(1+\frac{1}{\cos a}\right)\ge 3+2\sqrt{2}$? Prove the inequality:
$$\left(1+\dfrac{1}{\sin a}\right)\left(1+\dfrac{1}{\cos a}\right)\ge 3+2\sqrt{2}; \text{ for } a\in\left]0,\frac{\pi}{2}\right[$$
 A: 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 $$\left(1+\dfrac{1}{\sin a}\right)\left(1+\dfrac{1}{\cos a}\right) = 1+\dfrac{1}{\cos a}+\dfrac{1}{\sin a}+\dfrac{1}{\sin a\cos a} \\\geq 1+2\sqrt{\dfrac{1}{\sin a\cos a}}+\dfrac{1}{\sin a\cos a}\geq 1+2\sqrt{2} + 2 = 3+ 2\sqrt{2}$$
A: 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 $x=y$.
A: 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 $a = \frac{\pi}{4}$
A: 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 $(1+\sqrt{2})^2=1+2+2\sqrt{2}=3+2\sqrt{2}$
