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

Answer: 
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 $$\left(1+\dfrac{1}{\sin a}\right)\left(1+\dfrac{1}{\cos a}\right) = 1+\dfrac{1}{\cos a}+\dfrac{1}{\sin a}+\dfrac{1}{\sin a\cos a} \\\geq 1+2\sqrt{\dfrac{1}{\sin a\cos a}}+\dfrac{1}{\sin a\cos a}\geq 1+2\sqrt{2} + 2 = 3+ 2\sqrt{2}$$
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 $x=y$.