Trigonometric inequality in sec(x) and csc(x) How can I prove the following inequality
\begin{equation*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right) \geq 3+%
\sqrt{2},~~~\forall x\in \left( 0,\frac{\pi }{2}\right) .
\end{equation*}%
I tried the following
\begin{eqnarray*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right)  &\geq
&\left( 1+1\right) \left( 1+\frac{1}{\cos x}\right)  \\
&=&2\left( 1+\frac{1}{\cos x}\right)  \\
&\geq &2\left( 1+1\right) =4,
\end{eqnarray*}
but $4\leq 3+\sqrt{2}$.

Expand the expression to get
$$\left(1+\frac{1}{\sin x}\right)\left(1+\frac{1}{\cos x}\right)=1+\frac{1}{\sin x}+\frac{1}{\cos x}+\frac{1}{\sin x\cos x}$$
Then using the identity $\sin x\cos x = \frac{1}{2}\sin 2x$, rewrite as
\begin{eqnarray*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right)  &=
& 1+\frac{1}{\sin x}+\frac{1}{\cos x}+\frac{2}{\sin 2x} \\
&\geq&1+1+1+2 \text{ for }x\in (0,\frac{\pi}{2}) \\
&= &5\\
&\geq & 3+\sqrt{2}\text{ since } 2\geq \sqrt{2}
\end{eqnarray*}
