Finding the range of $\frac{1+\sin^4 x}{\sin^4 x}\cdot \frac{1+\cos^4 x}{\cos^4 x}$ 
Range of function $$f(x) = \frac{(1+\sin^4 x)}{\sin^4 x}\cdot \frac{(1+\cos^4 x)}{\cos^4 x}$$ 

Using $\bf{A.\geq G.M}$ Inequality,
$$1+\sin^4 x\geq 2\sin^2 x \qquad\text{and}\qquad 1+\cos^4 x\geq 2\cos^2 x$$
So $$\frac{(1+\sin^4 x)}{\sin^4 x}\cdot \frac{(1+\cos^4 x)}{\cos^4 x}\geq \frac{4}{\sin^2 x\cdot \cos^2 x} = \frac{16}{(\sin 2x)^2}\geq 16$$
But the answer  given as $\left[25,\infty\right)$.
Please help me. How can I solve it using an inequality or any other way. Thanks.
 A: Let $a=\sin^2 x, b = \cos^2 x$. Then $a+b=1$ and hence $\sqrt{ab} \leq \frac{a+b}{2} = \frac{1}{2}$. Thus $\frac{1}{ab} \geq 4$. Now the given expression is 
\begin{align*}
\left(\frac{1+a^2}{a^2}\right)\left(\frac{1+b^2}{b^2}\right) &= 1+\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{a^2b^2} \\
&\geq 1 + \frac{2}{ab} + \frac{1}{a^2b^2} \\
&\geq 1 + 8 + 16 = 25
\end{align*}
A: Answering to the question of why the OP method gives a lower bound for the codomain, consider (e.g.) the following:
$$
\begin{gathered}
  \left( {z^{\,2}  + 1} \right) \geqslant 2z\quad \left( {y^{\,2}  + 1} \right) \geqslant 2y\quad  \Rightarrow \quad \left( {z^{\,2}  + 1} \right)\left( {y^{\,2}  + 1} \right) \geqslant 4\,z\,y\quad  \Rightarrow  \hfill \\
   \Rightarrow \quad \left( {z^{\,2}  + 1} \right)\left( {y^{\,2}  + 1} \right) = 4\quad \left| {\;\left| z \right| = 1\; \wedge \;\left| y \right| = 1} \right. \hfill \\ 
\end{gathered} 
$$
Here, since each single inequality bears the $ \geqslant$ sign, implies that the equality is attained for some values of the variable.
When multiplied together , keeping the variables distinct , then surely for all
the $(y,z)$ combining the above values, the equality will be attained.
However, when in it we place $(y(x), z(x))$ we are going to restrict the values of
$y$ and $z$ to stay on the curve parametrized in $x$, which might not allow
$y$ and $z$ to attain "contemporarily" the equality (allow me to say that you might have $( = )\, \times \, ( > )$).
In particular, the substitution 
$$
\left\{ \begin{gathered}
  y = 1/\sin ^{\,2} x \hfill \\
  z = 1/\cos ^{\,2} x \hfill \\ 
\end{gathered}  \right.
$$
will produce the situation which is well evidentiated in the picture.  

