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 $4\leq 3+\sqrt{2}$.
 A: 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*}
A: 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.
A: 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 $\dfrac{\sin x+1}{\sin x}\cdot\dfrac{\cos x +1}{\cos x}\geq 3+2\sqrt{2}$.
A: Let $\left( 1+\dfrac1{\sin x}\right) \left( 1+\dfrac1{\cos x}\right)=1+y$
$\iff\sec x+\csc x=y-\sec x\csc x$
Squaring both sides, $$\sec^2x+\csc^2x+2\sec x\csc x=y^2-2y\sec x\csc x+\sec^2x\csc^2x$$
But $\sec^2x+\csc^2x=\dfrac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}=\sec^2x\csc^2x$
$$\implies y^2-(2y-1)\sec x\csc x=0\iff0=\sin2x y^2-4y-4\le y^2-4y-4$$ as $\sin2x\le1$
As the roots of $y^2-4y-4=0$ are $y=2\pm2\sqrt2$
Either $y\ge2+2\sqrt2$ or $y\le2-2\sqrt2$
But as $x\in \left( 0,\frac{\pi }{2}\right),y>0\implies y\not\le0>2-2\sqrt2$
Can you take it from here?
The equality occurs if $\sin2x=1\iff x=\dfrac\pi4(?)$
