<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] Let $$y=\left(1+\dfrac{1+t^2}{1-t^2}\right)\left(1+\dfrac{1+t^2}{2t}\right)=\dfrac{2(1+t)^2}{2t(1-t^2)}=\dfrac{1+t}{t(1-t)}$$  as $t+1\ne0$
$$\implies yt^2-t(y-1)+1=0$$
As $t$ is real, the discriminant $$(y-1)^2-4y\ge0$$
As the roots of $y^2-6y+1=0$ are $y=\dfrac{6\pm\sqrt{32}}2=3\pm2\sqrt2$
$y^2-6y+1\ge0\implies$ either $y\ge3+2\sqrt2$ or $y\le3-2\sqrt2$
As $0< x<\dfrac\pi2, \csc x,\sec x\ge1\implies y>3\implies y\not\le3-2\sqrt2$
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