<s>[INST] 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 $4\leq 3+\sqrt{2}$.
 [/INST] 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 $\sin2x=1\iff x=\dfrac\pi4(?)$
</s>