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?
Solution 1:
$$\sin^2\theta+\cos^2\theta+2+2+\csc^2\theta+\sec^2\theta$$
$$=5+\frac1{\sin^2\theta\cos^2\theta}=5+\frac4{(\sin2\theta)^2}=5+4\csc^22\theta$$
Now, $\csc^22\theta=1+\cot^22\theta\ge1$ for real $\theta$
The equality occurs if $\csc^22\theta=1\iff\sin^22\theta=1$
$\iff\cos2\theta=0\implies2\theta=(2n+1)\dfrac\pi2$ where $n$ is any integer
Solution 2:
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 $$\dfrac{1}{\sin^2{\theta}}+\dfrac{1}{\cos^2{\theta}}\ge\dfrac{(1+1)^2}{\sin^2{\theta}+\cos^2{\theta}}=4$$