<s>[INST] Prove $(1+\sec x)(1+\csc x) > 5$ if $x \in {\left]0,\frac{\pi}{2}\right[}$ 
Prove $(1+\sec x)(1+\csc x) > 5$ if $x \in  {\left]0,\frac{\pi}{2}\right[}$.

I have done the following:
Let $t=\tan \frac{x}{2}\in ]0,1[$, then using the $t$-formulae the LHS becomes
$$(1+\sec x)(1+\csc x) = \frac{2t+1+t^2}{t(1-t^2)}$$
I could simply it further but then I keep lowering the bound, now I continue as follows:
$$\frac{2t+1+t^2}{t(1-t^2)} = \frac{2}{1-t^2} + \frac{1}{t(1-t^2)}+\frac{t^2-1}{t(1-t^2)}+\frac{1}{t(1-t^2)}$$
Now since $1-t^2$ and $t \in ]0,1[$ then $\frac{1}{t(1-t^2)}> 1$ and so is $\frac{1}{1-t^2}$ then:
$$> 2+1+1-\frac{1}{t}>4$$
Close, but no cigar. Where could I improve the bounding?
 [/INST] $$(1+\sec x)(1+\csc x) = 1+\sec x+\csc x+\frac{1}{\sin(x)\cos(x)}$$
where $f(x)=\sec x+\csc x$ is a convex function over $\left(0,\frac{\pi}{2}\right)$, with the property that $f(x)=f\left(\frac{\pi}{2}-x\right)$. It follows that $x\in\left(0,\frac{\pi}{2}\right)$ implies $f(x)\geq f\left(\frac{\pi}{4}\right)$. The same holds for $g(x)=\frac{1}{\sin(x)\cos(x)}$, hence it follows that:
$$ \begin{eqnarray*}x\in\left(0,\frac{\pi}{2}\right)\quad\Longrightarrow\quad (1+\sec x)(1+\csc x)&\geq& 1+ f\left(\frac{\pi}{4}\right)+g\left(\frac{\pi}{4}\right)\\&=&3+2\sqrt{2}\\&>&5.\end{eqnarray*} $$
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