\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Notice that \begin {align*} \LHS & = \frac {1}{2} \left (I_{n + 1} + I_{n - 1}\right ) \\ & = \frac {1}{2} \left (\int _{0}^{\beta } (\sec x + \tan x)^{n + 1} \Diff x + \int _{0}^{\beta } (\sec x + \tan x)^{n - 1} \Diff x\right ) \\ & = \frac {1}{2} \int _{0}^{\beta } (\sec x + \tan x)^{n - 1} \left [(\sec x + \tan x)^2 + 1\right ] \Diff x \\ & = \frac {1}{2} \int _{0}^{\beta } (\sec x + \tan x)^{n - 1} \left (\sec ^2 x + \tan ^2 x + 2 \sec x \tan x + 1\right ) \Diff x \\ & = \frac {1}{2} \int _{0}^{\beta } (\sec x + \tan x)^{n - 1} \cdot 2\left (\sec ^2 x + \sec x \tan x\right ) \Diff x \\ & = \int _{0}^{\beta } (\sec x + \tan x)^{n - 1} \Diff (\sec x + \tan x) \\ & = \frac {1}{n} \left [\left (\sec x + \tan x\right )^{n}\right ]_{0}^{\beta } \\ & = \frac {1}{n} \left (\left (\sec \beta + \tan \beta \right )^{n} - \left (\sec 0 + \tan 0\right )^{n}\right ) \\ & = \frac {1}{n} \left (\left (\sec \beta + \tan \beta \right )^{n} - 1 \right ) \\ & = \RHS , \end {align*}
as desired.
To show the final part, we would like to show that \[ I_n < \frac {1}{2} \left (I_{n + 1} + I_{n - 1}\right ) = \frac {1}{n} \left (\left (\sec \beta + \tan \beta \right )^{n} - 1 \right ), \] which is equivalent to showing \[ I_{n + 1} + I_{n - 1} - 2 I_n > 0. \]
\begin {align*} & \phantom {=} I_{n + 1} + I_{n - 1} - 2 I_n \\ & = \int _{0}^{\beta } (\sec x + \tan x)^{n + 1} \Diff x + \int _{0}^{\beta } (\sec x + \tan x)^{n - 1} \Diff x - 2 \int _{0}^{\beta } (\sec x + \tan x)^{n} \Diff x \\ & = \int _{0}^{\beta } \left (\sec x + \tan x\right )^{n - 1} \left (2 \sec ^2 x + 2 \sec x \tan x - 2 \sec x - 2 \tan x\right ) \\ & = \int _{0}^{\beta } \left (\sec x + \tan x\right )^{n - 1} \left (\sec ^2 x + \tan ^2 x + 2 \sec x \tan x - 2 \sec x - 2 \tan x + 1 \right ) \\ & = \int _{0}^{\beta } \left (\sec x + \tan x\right )^{n - 1} \left [\left (\sec x + \tan x\right )^2 - 2 \left (\sec x + 2 \tan x\right ) + 1\right ] \\ & = \int _{0}^{\beta } \left (\sec x + \tan x\right )^{n - 1} \left (\left (\sec x + \tan x\right ) - 1\right )^2. \end {align*}
For \(0 \leq x < \frac {\pi }{2}\), \(\sec x > 0, \tan x > 0\), and so \(\sec x + \tan x > 0\), \(\left (\sec x + \tan x\right )^{n - 1} > 0\).
Also, \(\sec x = \frac {1}{\cos x} > \frac {1}{1} = 1\), and hence \(\sec x + \tan x - 1 > 0\), so \(\left (\left (\sec x + \tan x\right ) - 1\right )^2 > 0\).
Hence, the integrand is greater than \(0\) on \((0, \beta ) \subseteq \left (0, \frac {\pi }{2}\right )\).
This shows that the desired equation is greater than \(0\), and hence, we have the desired inequality as desired.
Notice that \begin {align*} \frac {1}{2} \left (J_{n + 1} + J_{n - 1}\right ) & = \frac {1}{2} \int _{0}^{\beta } (\sec x \cos \beta + \tan x)^{n - 1} \left [\left (\sec x \cos \beta + \tan x\right )^2 + 1\right ] \Diff x \\ & = \frac {1}{2} \int _{0}^{\beta } (\sec x \cos \beta + \tan x)^{n - 1} \left [\sec ^2 x \cos ^2 \beta + \tan ^2 x + 2 \sec x \tan x \cos \beta + 1\right ] \Diff x \\ & = \frac {1}{2} \int _{0}^{\beta } (\sec x \cos \beta + \tan x)^{n - 1} \left (\sec ^2 x \cos ^2 \beta + \sec ^2 x + 2 \sec x \tan x \cos \beta \right ) \Diff x \\ & = \frac {1}{2} \int _{0}^{\beta } (\sec x \cos \beta + \tan x)^{n - 1} \left (2 \sec ^2 x - \sec ^2 x \sin ^2 \beta + 2 \sec x \tan x \cos \beta \right ) \Diff x \\ & = \int _{0}^{\beta } (\sec x \cos \beta + \tan x)^{n - 1} \left (\sec ^2 x + \sec x \tan x \cos \beta \right ) \Diff x \\ & \phantom {=} - \frac {\sin ^2 \beta }{2} \int _{0}^{\beta } (\sec x \cos \beta + \tan x)^{n - 1} \sec ^2 x \Diff x. \end {align*}
The first part of the integral integrates similarly: \begin {align*} & \phantom {=} \int _{0}^{\beta } (\sec x \cos \beta + \tan x)^{n - 1} \left (\sec ^2 x + \sec x \tan x \cos \beta \right ) \Diff x \\ & = \int _{0}^{\beta } (\sec x \cos \beta + \tan x)^{n - 1} \Diff \left (\sec x \cos \beta + \tan x\right ) \\ & = \frac {1}{n} \left [\left (\sec x \cos \beta + \tan x\right )^{n}\right ]_{0}^{\beta } \\ & = \frac {1}{n} \left [\left (\sec \beta \cos \beta + \tan \beta \right )^n - \left (\sec 0 \cos \beta + \tan 0\right )^n\right ] \\ & = \frac {1}{n} \left [(1 + \tan \beta )^n - \cos ^n \beta \right ]. \end {align*}
The second part of the integral has a positive integrand over \((0, \beta )\), and hence the integral is positive, which means \begin {align*} \frac {1}{2} \left (J_{n + 1} + J_{n - 1}\right ) & > \int _{0}^{\beta } (\sec x \cos \beta + \tan x)^{n - 1} \left (\sec ^2 x + \sec x \tan x \cos \beta \right ) \Diff x \\ & = \frac {1}{n} \left [(1 + \tan \beta )^n - \cos ^n \beta \right ]. \end {align*}
We would like to show that \(J_{n + 1} + J_{n - 1} - 2 J_{n} > 0\) similar as before to show the final result. Note that \begin {align*} & \phantom {=} J_{n + 1} + J_{n - 1} - 2 J_{n} \\ & = \int _{0}^{\beta } \left (\sec x \cos \beta + \tan x\right )^{n - 1} \left [\left (\sec x \cos \beta + \tan x\right )^2 + 1 - 2 \left (\sec x \cos \beta + \tan x\right )\right ] \Diff x \\ & = \int _{0}^{\beta } \left (\sec x \cos \beta + \tan x\right )^{n - 1} \left [(\sec x \cos \beta + \tan x) - 1\right ]^2 \Diff x \\ & > 0, \end {align*}
and hence \(J_{n} < \frac {1}{2} \left (J_{n + 1} + J_{n - 1}\right )\), which shows \[ J_{n} < \frac {1}{n} \left ((1 + \tan \beta )^{n} - \cos ^{n} \beta \right ), \] as desired.