We shall prove that if $f(x)=\left(1+\dfrac{1}{\sin x}\right)\left(1+\dfrac{1}{\cos x}\right)$ and $g(x)=5\left[1+x^4\left(\dfrac{\pi}{2}-x\right)^4\right]$, $0<x<\dfrac{\pi}{2}$, then
$\text{min} f(x)>5.8>\text{max} g(x)$
Since $f(x)$ is symmetric about the point $x=\dfrac{\pi}{4}$ in $\left(0,\,\dfrac{\pi}{2}\right)$, we may use the substitution $x=\dfrac{\pi}{4}-t$, where $-\dfrac{\pi}{4}<t<\dfrac{\pi}{4}$, then
$\begin{align*}f(x)&=\left(1+\dfrac{1}{\sin\left(\dfrac{\pi}{4}-t\right)}\right)\left(1+\dfrac{1}{\cos\left(\dfrac{\pi}{4}-t\right)}\right)\\&=\dfrac{\left(\dfrac{1}{\sqrt{2}}(\cos t -\sin t)+1\right)\left(\dfrac{1}{\sqrt{2}}(\cos t +\sin t)+1\right)}{\sin \left(\dfrac{\pi}{4}-t\right)\cos \left(\dfrac{\pi}{4}-t\right)}\\&=\dfrac{(\sqrt{2}+\cos t-\sin t)(\sqrt{2}+\cos t+\sin t)}{2\sin \left(\dfrac{\pi}{4}-t\right)\cos \left(\dfrac{\pi}{4}-t\right)}\\&=\dfrac{(\sqrt{2}+\cos t)^2-\sin^2 t}{\sin \left(\dfrac{\pi}{4}-t\right)}\\&=\dfrac{2+2\sqrt{2}\cos t+\cos 2t}{\cos 2t}\\&=1+\dfrac{2(\sqrt{2}\cos t+1)}{2\cos^2 t-1}\\&=1+\dfrac{2}{\sqrt{2}\cos t-1}\end{align*}$
For $f(x)$ to be at a minimum, $\sqrt{2}\cos t-1$ is at a maximum and so $\cos t=1$. This happens for $t=0$, that is, $x=\dfrac{\pi}{4}$. Thus,
$\text{min} f(x)=1+\dfrac{2}{\sqrt{2}-1}=3+2\sqrt{2}>3+2(1.4)=5.8$.
Now, the maximum of $x\left(\dfrac{\pi}{2}-x\right)$ is $\dfrac{\pi^2}{16}$, which is attained at $x=\dfrac{\pi}{4}$, as
$x\left(\dfrac{\pi}{2}-x\right)=\dfrac{\pi^2}{16}-\left(\dfrac{\pi}{4}-x\right)^2$
So
$\text{max} g(x)=5+\left(\dfrac{\pi^2}{16}\right)^4=5+\dfrac{\pi^8}{16^4}$
Since $\pi^2<10$ we see that
$\begin{align*}\text{max} g(x)&<5\left(1+\dfrac{10^4}{16^4}\right)\\&=5\left(1+\dfrac{10^6}{16^4\times 100}\right)\\&=5\left(1+\dfrac{(10^3)^2}{2^{16}\times 100}\right)\\&<5\left(1+\dfrac{(2^{10})^2}{2^{16}\times 100}\right)\\&=5\left(1+\dfrac{2^4}{100}\right)\\&=5(1+0.16)\\&=5.8\end{align*}$
Hence the inequality follows.
