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?
 A: 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
$$\dfrac{1}{\sin^2{\theta}}+\dfrac{1}{\cos^2{\theta}}\ge\dfrac{(1+1)^2}{\sin^2{\theta}+\cos^2{\theta}}=4$$
A: $\begin{align}(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 & =5+\sec^2 x+\csc^2 x\\&=5+1+\tan^2 x+1+\cot^2x\\&=7+\tan^2x+\cot^2x\\&\geq9\end{align}$ 
A: $$\sin^2\theta+\cos^2\theta+2+2+\csc^2\theta+\sec^2\theta$$
$$=5+\frac1{\sin^2\theta\cos^2\theta}=5+\frac4{(\sin2\theta)^2}=5+4\csc^22\theta$$
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
A: 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.
A: Using Cauchy–Schwarz inequality:
$$(\sin x + \csc x)^2+(\cos x + \sec x)^2\\=[(\sin x + \csc x)^2+(\cos x + \sec x)^2][\sin^2 x+ \cos^2 x]\\\geq \mid (\sin x + \csc x)\sin x+(\cos x + \sec x)\cos x \mid^2\\=\mid (\sin^2 x +1)+(\cos^2 x +1) \mid^2\\=\mid 1+1+1 \mid^2=9$$
