\(% 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}}\)
Since \(X \sim \Uniform [a, b]\), we must have for the probability density function of \(X\), that \[ f_{X}(x) = \frac {1}{b - a} \] for \(x \in [a, b]\), and \(0\) everywhere else. Hence, the cumulative distribution function of \(X\) is \[ F_{X}(x) = \begin {cases} 0, & x \leq a, \\ \frac {x - a}{b - a}, & a < x \leq b, \\ b, & \text {otherwise}. \end {cases} \]
Since \(f\) is bijective and strictly decreasing on \([a, b]\), we must have for \(y \in [a, b]\), that \begin {align*} \Prob (Y \leq y) & = \Prob (f(X) \leq y) \\ & = \Prob (X \geq f^{-1}(y)) \\ & = \Prob (X \geq f(y)) \\ & = 1 - \Prob (X < f(y)) \\ & = 1 - F_{X} (f(y)) \\ & = 1 - \frac {f(y) - a}{b - a} \\ & = \frac {(b - a) - (f(y) - a)}{b - a} \\ & = \frac {b - f(y)}{b - a}, \end {align*}
as desired.
Hence, by differentiation with respect to \(y\), we have the probability density function of \(Y\) satisfies \[ f_{Y}(y) = - \frac {f'(y)}{b - a}. \]
Hence, by the definition of expectation, we have \begin {align*} \Expt (y^2) & = \int _{a}^{b} f_{Y}(y) y^2 \Diff y \\ & = -\frac {1}{b - a} \int _{a}^{b} - f'(y) y^2 \Diff y \\ & = -\frac {1}{b - a} \int _{a}^{b} y^2 \Diff f(y) \\ & = \frac {1}{b - a} \left [-\left [y^2 f(y)\right ]_{a}^{b} - 2\int _{a}^{b} y f(y) \Diff Y\right ] \\ & = \frac {1}{b - a} \left [- b^2 f(b) + a^2 f(a) + 2\int _{a}^{b} y f(y) \Diff Y\right ] \\ & = \frac {1}{b - a} \left [\frac {b}{3} (b^3 - a^3) - b^2a + a^2b + 2\int _{a}^{b} y f(y) \Diff x\right ] \\ & = \frac {b}{3} \left (b^2 + ab + a^2\right ) - ab + \int _{a}^{b} \frac {2xf(x) \Diff x}{b - a}. \end {align*}
Since \(\frac {1}{Z} + \frac {1}{X} = \frac {1}{c}\), by rearranging, we have \[ Z = \frac {1}{\frac {1}{c} - \frac {1}{X}} = \frac {cX}{X - c}. \]
By given, we have \[ c = \frac {ab}{a + b}, \] and hence \[ c < a, c < b. \]
Let \(f(x) = \frac {cx}{x - c}\). Notice that \begin {align*} f(a) & = \frac {ac}{a - c} \\ & = \frac {a^2b / (a + b)}{a - ab / (a + b)} \\ & = \frac {a^2 b}{a^2 + ab - ab} \\ & = b, \end {align*}
and \begin {align*} f(b) & = \frac {bc}{b - c} \\ & = \frac {ab^2 / (a + b)}{b - ab / (a + b)} \\ & = \frac {ab^2}{b^2 + ab - ab} \\ & = a. \end {align*}
Also, since \[ f(x) = \frac {1}{\frac {x - c}{cx}} = \frac {1}{\frac {1}{c} - \frac {1}{x}}, \] as \(x\) strictly increases, \(\frac {1}{x}\) strictly decreases, \(- \frac {1}{x}\) strictly increases, the denominator strictly increases, and hence \(f(x)\) strictly decreases.
Note that \[ \frac {1}{f(x)} + \frac {1}{x} = \frac {1}{c}, \] and hence \[ \frac {1}{x} + \frac {1}{f^{-1}(x)} = \frac {1}{c}, \] which implies \[ f(x) = f^{-1}(x). \]
So \(Z = f(X)\) for this \(f\) satisfying all three conditions above. Hence, \begin {align*} \Expt (Z) & = \int _{a}^{b} f(x) f_{X} (x) \Diff x \\ & = \frac {1}{b - a} \int _{a}^{b} \frac {cx \Diff x}{x - c} \\ & = \frac {1}{b - a} \int _{a}^{b} \left (c + \frac {c^2}{x - c}\right ) \Diff x \\ & = \frac {1}{b - a} \left [cx + c^2 \ln \abs *{x - c}\right ]_{a}^{b} \\ & = \frac {1}{b - a} \left [\left (cb + c^2 \ln \abs *{b - c}\right ) - \left (ca + c^2 \ln \abs *{a - c}\right )\right ] \\ & = c + \frac {c^2}{b - a} \ln \abs *{\frac {b - c}{a - c}} \\ & = c + \frac {c^2}{b - a} \ln \left (\frac {b - c}{a - c}\right ), \end {align*}
and using the result from the previous part, \begin {align*} \Expt (Z^2) & = -ab + \int _{a}^{b} \frac {2x f(x)}{b - a} \Diff x \\ & = -ab + \frac {2}{b - a} \cdot \int _{a}^{b} \frac {cx^2}{x - c} \Diff x \\ & = -ab + \frac {2c}{b - a} \cdot \int _{a}^{b} \left (x + c + \frac {c^2}{x - c}\right ) \Diff x \\ & = -ab + \frac {2c}{b - a} \cdot \left [\frac {x^2}{2} + cx + c^2 \ln \abs *{x - c}\right ]_{a}^{b} \\ & = -ab + \frac {2c}{b - a} \cdot \left [\left (\frac {b^2}{2} + bc + c^2 \ln \abs *{b - c}\right ) - \left (\frac {a^2}{2} + ac + c^2 \ln \abs *{a - c}\right )\right ] \\ & = -ab + \frac {2c}{b - a} \cdot \left [(b - a) \left (c + \frac {a + b}{2}\right ) + c^2 \ln \abs *{\frac {b - c}{a - c}}\right ] \\ & = -ab + 2c \left (c + \frac {a + b}{2}\right ) + \frac {2c^3}{b - a} \ln \left (\frac {b - c}{a - c}\right ) \\ & = 2c^2 + (a + b)c - ab + \frac {2c^3}{b - a} \ln \left (\frac {b - c}{a - c}\right ) \\ & = 2c^2 + \frac {2c^3}{b - a} \ln \left (\frac {b - c}{a - c}\right ). \end {align*}
Hence, the variance of \(Z\) satisfies that \begin {align*} \Var (Z) & = \Expt (Z^2) - \Expt (Z)^2 \\ & = 2c^2 + \frac {2c^3}{b - a} \ln \left (\frac {b - c}{a - c}\right ) - \left (c + \frac {c^2}{b - a} \ln \left (\frac {b - c}{a - c}\right )\right ) \\ & = 2c^2 + \frac {2c^3}{b - a} \ln \left (\frac {b - c}{a - c}\right ) - c^2 - \frac {2c^3}{b - a} \ln \left (\frac {b - c}{a - c}\right ) - \frac {c^4}{(b - a)^2} \ln \left (\frac {b - c}{a - c}\right )^2 \\ & = c^2 - \frac {c^4}{(b - a)^2} \ln \left (\frac {b - c}{a - c}\right )^2. \end {align*}
Therefore, since the variance of a non-constant random variable is always positive, \begin {align*} c^2 & > \frac {c^4}{(b - a)^2} \ln \left (\frac {b - c}{a - c}\right )^2 \\ (b - a)^2 & > c^2 \ln \left (\frac {b - c}{a - c}\right )^2 \\ \abs *{b - a} & > \abs *{c \ln \left (\frac {b - c}{a - c}\right )} \\ \abs *{\ln \left (\frac {b - c}{a - c}\right )} & < \abs *{\frac {b - a}{c}}. \end {align*}
Notice that since \(b > a\), we must have \(b - c > a - c\), so the natural log on the left-hand side is positive, and the fraction within the absolute value on the right-hand side is positive as well, and hence \[ \ln \left (\frac {b - c}{a - c}\right ) < \frac {b - a}{c}. \]