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

If we set $t=\tan(x/2)$, we know that $0<t<1$ and the inequality is
$$
\left(1+\frac{1+t^2}{2t}\right)
\left(1+\frac{1+t^2}{1-t^2}\right)
\ge 3+\sqrt{2}
$$
or
$$
\frac{(1+t)^2}{2t}\frac{2}{1-t^2}\ge3+\sqrt{2}
$$
that is
$$
\frac{1+t}{t(1-t)}\ge 3+\sqrt{2}
$$
that becomes (with the condition $0<t<1$)
$$
(3+\sqrt{2})t^2-(2+\sqrt{2})t+1\ge0
$$
which has negative discriminant, so it's true.

If we consider the function, defined on $(0,1)$,
$$
f(t)=\frac{1+t}{t-t^2}
$$
we have
$$
f'(t)=\frac{t-t^2-(1+t)(1-2t)}{(t-t^2)^2}=
\frac{t^2+2t-1}{(t-t^2)^2}
$$
that vanishes at $t=\sqrt{2}-1$, which is the point of minimum. We can compute
$$
f(\sqrt{2}-1)=3+2\sqrt{2}
$$
so this is the best bound.
