<s>[INST] To prove $(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 \ge 9$ I used the following way but got wrong answer
$$A.M. \ge  G.M.$$
$$ \frac{\sin \theta + \csc  \theta}{2} \ge \sqrt{\sin \theta \cdot \csc \theta}$$
Squaring both sides,
\begin{equation*}
(\sin\theta + \csc\theta )^2 \ge 4 \tag{1} 
\end{equation*}
Similarly
\begin{equation*}
(\cos\theta + \sec\theta )^2 \ge 4 \tag{2}
\end{equation*}
Adding equation (1)  and (2)
\begin{equation*}
(\sin\theta + \csc\theta )^2+(\cos\theta + \sec\theta )^2 \ge 8
\end{equation*}
What is wrong?
 [/INST] Without AM-GM, you could expand the expression which, after simplifications, write $$f(x)=(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 =\csc ^2(x)+\sec ^2(x)+5$$ The derivative write $$f'(x)=2 \tan (x) \sec ^2(x)-2 \cot (x) \csc ^2(x)=-8 \sin (4 x) \csc ^4(2 x)$$ and cancels for $x=\frac{\pi}{4}, \frac{3\pi}{4}, \cdots$ and, for these values, $f(x)=9$. The second derivative test shows that this is a minimum value.
</s>