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 $4\leq 3+\sqrt{2}$.

We have $$\dfrac{\sin x+1}{\sin x}\cdot\dfrac{\cos x +1}{\cos x}=\dfrac{\sin x \cos x+\sin x +\cos x +1}{\sin x \cos x}=\dfrac{1/2 \sin 2x+\sqrt{2}\sin(\frac{\pi}{4}+x)+1}{1/2\sin 2x}$$
$=1+\dfrac{2\sqrt{2}\sin(\frac{\pi}{4}+x)+2}{\sin 2x}$ is at its minimum when $x=\pi/4$ since we want the denominator closest to 1.
At $x=\pi/4$ the expression becomes $1+2\sqrt{2}+2=3+2\sqrt{2}$
Thus, we have shown the inequality $\dfrac{\sin x+1}{\sin x}\cdot\dfrac{\cos x +1}{\cos x}\geq 3+2\sqrt{2}$.
