<s>[INST] 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.
 [/INST] 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*}
</s>