Problem 949

Prove that from equation (4), taking $p=-\frac{n}{2}<0$, we have
$$\begin{aligned} y & =\frac{\left(\sec ^{2} \alpha\right)^{p}}{\left(a^{\frac{1}{1-p}}\right)^{p-1}}-\frac{\left(\tan ^{2} \alpha\right)^{p}}{\left(b^{\frac{1}{1-p}}\right)^{p-1}} \\ & \leqslant \frac{\left(\sec ^{2} \alpha-\tan ^{2} \alpha\right)^{p}}{\left(a^{\frac{1}{1-p}}-b^{\frac{1}{1-p}}\right)^{p-1}} \\ & =\left(a^{\frac{2}{n+2}}-b^{\frac{2}{n+2}}\right)^{\frac{n+2}{2}} . \end{aligned}$$

Similarly, we have
$$\begin{array}{l} \text { (1) } y=a \sec ^{n} \alpha-b \tan ^{n} \alpha \\ \leqslant\left(a^{\frac{2}{2-n}}-b^{\frac{2}{2-n}}\right)^{\frac{2-n}{2}}\left(p=\frac{n}{2}\right) ; \\ \text { (2) } y=\frac{a}{\sqrt[n]{\sec \alpha}}-\frac{b}{\sqrt[n]{\tan \alpha}} \\ \leqslant\left(a^{\frac{2 n}{2 n+1}}-b^{\frac{2 n}{2 n+1}}\right)^{\frac{2 n+1}{2 n}}\left(p=-\frac{1}{2 n}\right) ; \end{array}$$
(3) $y=a \sqrt[n]{\sec \alpha}-b \sqrt[n]{\tan \alpha}$
$$\geqslant\left(a^{\frac{2 n}{2 n-1}}-b^{\frac{2 n}{2 n-1}}\right)^{\frac{2 n-1}{2 n}}\left(p=\frac{1}{2 n}\right) \text {. }$$

It is not difficult to find that the condition for equality in the above inequalities is
$$\frac{\sec ^{2} \alpha}{a^{\frac{1}{1-p}}}=\frac{\tan ^{2} \alpha}{b^{\frac{1}{1-p}}}=\frac{1}{a^{\frac{1}{1-p}}-b^{\frac{1}{1-p}}},$$

i.e., $\tan ^{2} \alpha=\frac{b^{\frac{1}{1-\rho}}}{a^{\frac{1}{1-p}}-b^{\frac{1}{1-p}}}$.