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?

Since
$$\left(\sin{\theta}+\dfrac{1}{\sin{\theta}}\right)^2+\left(\cos{\theta}+\dfrac{1}{\cos{\theta}}\right)^2=5+\dfrac{1}{\sin^2{\theta}}+\dfrac{1}{\cos^2{\theta}}$$
Use Cauchy-Schwarz inequality we have
$$\dfrac{1}{\sin^2{\theta}}+\dfrac{1}{\cos^2{\theta}}\ge\dfrac{(1+1)^2}{\sin^2{\theta}+\cos^2{\theta}}=4$$
