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?
 A: $$(1+\sec x)(1+\csc x) = 1+\sec x+\csc x+\frac{1}{\sin(x)\cos(x)}$$
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:
$$ \begin{eqnarray*}x\in\left(0,\frac{\pi}{2}\right)\quad\Longrightarrow\quad (1+\sec x)(1+\csc x)&\geq& 1+ f\left(\frac{\pi}{4}\right)+g\left(\frac{\pi}{4}\right)\\&=&3+2\sqrt{2}\\&>&5.\end{eqnarray*} $$
A: You can see that your function has local minimum somewhere between 1 and pi/2. First of all calculate where it is.
The local minimum of (1 + 1/sin)(1 + 1/cos) is where its derivative is equal to 0.
Sin **3 + sin **2 - cos **3 - cos **2 = 0
Substitute sin(x) as t and cos(x) as q.
T **3 + t **2 = q **3 + q **2
T = q
Sin (x) = cos (x)
There is one solution in range from 0 to pi/2 and it is pi/4.
Than you just insert the minimum to your function.
(1 + 1/sin(pi/4))(1 + 1/cos(pi/4)) = ~5.8
(1 + 1/sin)(1 + 1/cos) is always greater than 5.8 when x is in range from 0 to pi/2.
I hope this is answer to your question.
Please excuse lack of tex in this post, i am currenty on my phone.
A: 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 $0< x<\dfrac\pi2, \csc x,\sec x\ge1\implies y>3\implies y\not\le3-2\sqrt2$
