To prove: $e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\gt e^2$ Question:
Statement I: If $0\lt\theta\lt\frac\pi2$, then $e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\gt e^2$
Statement II: AM of positive numbers is always greater than or equal to their GM.
A) Both the statements are true and Statement II is correct explanation of Statement I.
B) Both the Statements are true and Statement II is not the correct explanation of Statement I.
C) Statement I is true and Statement II is false.
D) Statement I is false and Statement II is true.
My Attempt: I can see that both the statements are true. I wonder how to prove Statement I.
Answer given is option A.
$\frac{e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta}2\ge\sqrt{e^{\csc^2\theta}\sin^2\theta e^{\sec^2\theta}\cos^2\theta}$
Thus, $$RHS=\sqrt{e^{\frac1{\sin^2\theta}+\frac1{\cos^2\theta}}\sin^2\theta\cos^2\theta}\\ =e^{\frac1{\sin\theta\cos\theta}}\sin\theta\cos\theta$$
Not able to conclude.
 A: Background:
Expanding on Anne Bauval's comment.
Weighted AM-GM inequality
Let the nonnegative numbers $x_1, x_2, . . . , x_n$ and the nonnegative weights $w_1, w_2, . . . , w_n$ be given. Set $w = w_1 + w_2 + · · · + w_n$. If $w > 0$, then
${\displaystyle {\frac {w_{1}x_{1}+w_{2}x_{2}+\cdots +w_{n}x_{n}}{w}}\geq {\sqrt[{w}]{x_{1}^{w_{1}}x_{2}^{w_{2}}\cdots x_{n}^{w_{n}}}}}$
Answer:
Let $w_1=\sin^2\theta, w_2=\cos^2\theta\implies w=w_1+w_2=1$
Let $x_1=e^{\frac1{\sin^2\theta}}, x_2=e^{\frac1{\cos^2\theta}}$, thus,
$$\frac{e^{\frac1{sin^2\theta}}\sin^2\theta+e^{\frac1{\cos^2\theta}}\cos^2\theta}1\ge (e^{\frac1{\sin^2\theta}})^{\sin^2\theta}(e^{\frac1{\cos^2\theta}})^{\cos^2\theta}$$
Thus, $e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\ge e^2$
A: The given answer is wrong. The inequality in the title is false. When $\theta =\frac {\pi } 4$ you get equality instead of strict inequality.
A: The inequality is false as stated. However, we can show that $$e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\geq e^2 \tag{1}$$
for $0 < \theta < \frac\pi 2$. Using the AM-GM inequality,
$$e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\geq 2 e^{\frac{1}{2\sin^2\theta \cos^2\theta}} \sin\theta\cos\theta = 2 \phi(\sin\theta\cos\theta)$$
where $$\phi(x):= xe^{\frac{1}{2x^2}}$$
for $-\frac 12 \le x\le \frac12$. Of course,
$$2\phi(\sin\theta\cos\theta) \ge 2\inf_{\theta\in \left(0, \frac\pi 2 \right)} \phi(\sin\theta\cos\theta)$$ for every $\theta\in \left(0, \frac\pi 2 \right)$.  To compute the infimum on the right-hand side, we shall analyze the behavior of $\phi(x)$ for $-\frac 12 \le x\le \frac12$ (since $-\frac 12 \le \sin\theta\cos\theta\le \frac12$). Differentiating $\phi$, we have
$$\phi'(x) = e^{\frac{1}{2x^2}}\left(1 - \frac{1}{x^2} \right)$$
Since $-\frac 12 \le x\le \frac12$, we have $-\infty < 1 - \frac{1}{x^2} \le -3$. Therefore, $\phi'(x) < 0$ for $-\frac 12 \le x\le \frac12$.
With this information, it is clear that $\phi$ is strictly decreasing, and so it attains its minimum value at $x = \frac12$, or $\theta = \frac{\pi}{4}$. Putting everything together, we obtain
$$e^{\csc^2\theta}\sin^2\theta+e^{\sec^2\theta}\cos^2\theta\geq 2 \phi(\sin\theta\cos\theta) \ge 2\phi\left(\frac12 \right) = 2 \cdot \frac12  \cdot e^{2} = e^2$$
as desired.
