<s>[INST] If $\frac{\sec^8 \theta}{a}+\frac{\tan^8 \theta}{b} = \frac{1}{a+b}\;,$ Then prove that $ab\leq 0$ 
If $\displaystyle \frac{\sec^8 \theta}{a}+\frac{\tan^8 \theta}{b} = \frac{1}{a+b}\;,$  Prove that $ab\leq 0$

$\bf{My\; Try::}$ I am Trying To solve it Using Inequality.
Using $\bf{Cauchy\; Schwartz\; Inequality::}$
$$\frac{(\sec^4 \theta)^2}{a}+\frac{(\tan^4 \theta)^2}{b}\geq \frac{(\sec^4 \theta+\tan^4\theta)^2}{a+b}$$
and equality hold when $$\frac{\sec^4 \theta}{a}=\frac{\tan^4 \theta}{b}$$
Now I did not understand How can I solve after that
Help me
Thanks
 [/INST] Trivially, $\sec^8 \theta \ge 1$ and $\tan^8 \theta \ge 0$ for all $\theta \in \mathbb{R}$ for which $\sec \theta$ and $\tan \theta$ are defined.
If $a > 0$ and $b > 0$. Then, $\dfrac{\sec^8 \theta}{a}+\dfrac{\tan^8 \theta}{b} \ge \dfrac{1}{a}+\dfrac{0}{b} = \dfrac{1}{a} > \dfrac{1}{a+b}$, a contradiction.
If $a < 0$ and $b < 0$. Then, $\dfrac{\sec^8 \theta}{a}+\dfrac{\tan^8 \theta}{b} \le \dfrac{1}{a}+\dfrac{0}{b} = \dfrac{1}{a} < \dfrac{1}{a+b}$, a contradiction.
Thus, either $a \ge 0$ and $b \le 0$ or $a \le 0$ and $b \ge 0$. Therefore, $ab \le 0$.
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