\(% 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}}\)
If \(f(x) = 1\), this gives \[ \left (\int _{a}^{b} g(x) \Diff x\right )^2 \leq (b - a) \left (\int _{a}^{b} g(x)^2 \Diff x\right ). \]
Let \(g(x) = e^x\), \(a = 0\) and \(b = t\), and we have \[ \LHS = \left (\int _{0}^{t} e^x \Diff x\right )^2 = \left (e^t - 1\right )^2, \] and \[ \RHS = t \int _{0}^{t} e^{2x} \Diff x = \frac {t}{2} \left (e^{2t} - 1\right ) = \frac {t}{2} \left (e^t - 1\right ) \left (e^t + 1\right ). \]
Since \(\LHS \leq \RHS \), we have \[ \left (e^t - 1\right )^2 \leq \frac {t}{2} \left (e^t - 1\right ) \left (e^t + 1\right ), \] and hence \[ \frac {e^t - 1}{e^t + 1} \leq \frac {t}{2}, \] since \(e^t + 1 > 0\).
If \(f(x) = x\), and \(a = 0, b = 1\), the Schwarz inequality gives \[ \left (\int _{0}^{1} x g(x) \Diff x\right )^2 \leq \int _{0}^{1} x^2 \Diff x \int _{0}^{1} g(x)^2 \Diff x. \]
Since \[ \int _{0}^{1} x^2 \Diff x = \frac {1}{3} \left [x^3\right ]_{0}^{1} = \frac {1}{3}, \] we therefore have \[ 3\left (\int _{0}^{1} x g(x) \Diff x\right )^2 \leq \int _{0}^{1} g(x)^2 \Diff x. \]
Consider \(g(x) = \exp \left (-\frac {1}{4} x^2\right )\). Notice that \begin {align*} \int _{0}^{1} x g(x) \Diff x & = \int _{0}^{1} x \exp \left (-\frac {1}{4} x^2\right ) \Diff x \\ & = -2 \left [\exp \left (-\frac {1}{4} x^2\right )\right ]_{0}^{1} \\ & = -2 \left [\exp \left (-\frac {1}{4}\right ) - \exp \left (0\right )\right ] \\ & = 2 \left (1 - \exp \left (-\frac {1}{4}\right )\right ), \end {align*}
and hence \[ 3 \cdot \left [2 \left (1 - \exp \left (-\frac {1}{4}\right )\right )\right ]^2 \leq \int _{0}^{1} \exp \left (-\frac {1}{2} x^2\right ) \Diff x, \] which is equivalent to \[ \int _{0}^{1} \exp \left (-\frac {1}{2} x^2\right ) \Diff x \geq 12 \left (1 - \exp \left (-\frac {1}{4}\right )\right )^2, \] as desired.
For the right-half of the inequality, let \(f(x) = 1\), and let the bounds be \(a = 0, b = \frac {1}{2}\pi \), we have \[ \left (\int _{0}^{\frac {\pi }{2}} g(x) \Diff x\right )^2 \leq \frac {\pi }{2} \int _{0}^{\frac {\pi }{2}} g(x)^2 \Diff x. \]
Let \(g(x) = \sqrt {\sin x}\), and hence \[ \left (\int _{0}^{\frac {\pi }{2}} \sqrt {\sin x} \Diff x\right )^2 \leq \frac {\pi }{2} \int _{0}^{\frac {\pi }{2}} \sin x \Diff x = \frac {\pi }{2} \left [- \cos x\right ]_{0}^{\frac {\pi }{2}} = \frac {\pi }{2}. \]
Since the integrand \(\sqrt {\sin x} \geq 0\) for all \(x \in \left [0, \frac {\pi }{2}\right ]\), the integral over this interval must be non-negative, and hence \[ \int _{0}^{\frac {\pi }{2}} \sqrt {\sin x} \Diff x \leq \sqrt {\frac {\pi }{2}}. \]
For the left-half of the inequality, consider \(g(x) = \sqrt [4]{\sin x}\), and \(f(x) = \cos x\) (with the same bounds, \(a = 0, b = \frac {1}{2}\pi \)). We notice that \begin {align*} \int _{a}^{b} f(x) g(x) \Diff x & = \int _{0}^{\frac {1}{2} \pi } \cos x \sqrt [4] {\sin x} \Diff x \\ & = \frac {4}{5} \left [\left (\sin x\right )^{\frac {5}{4}}\right ]_{0}^{\frac {1}{2}} \\ & = \frac {4}{5} \left [1^{\frac {5}{4}} - 0^{\frac {5}{4}}\right ] \\ & = \frac {4}{5}, \end {align*}
and that \begin {align*} \int _{a}^{b} f(x)^2 \Diff x & = \int _{0}^{\frac {1}{2} \pi } \cos ^2 x \Diff x \\ & = \int _{0}^{\frac {1}{2} \pi } \frac {1 + \cos 2x}{2} \Diff x \\ & = \left [\frac {1}{2}x + \frac {1}{4} \sin 2x\right ]_{0}^{\frac {1}{2}\pi } \\ & = \left [\frac {1}{2} \cdot \frac {1}{2} \pi + \frac {1}{4} \sin \pi \right ] - \left [\frac {1}{2} \cdot 0 + \frac {1}{4} \sin 0\right ] \\ & = \frac {1}{4} \pi . \end {align*}
Hence, by the Schwarz inequality, we have \begin {align*} \frac {16}{25} \leq \frac {1}{4} \pi \cdot \int _{0}^{\frac {\pi }{2}} \sqrt {\sin x} \Diff x, \end {align*}
and hence \[ \int _{0}^{\frac {\pi }{2}} \sqrt {\sin x} \Diff x \geq \frac {64}{25\pi }. \]
Combining both sides of the equality, we hence have \[ \frac {64}{25 \pi } \leq \int _{0}^{\frac {\pi }{2}} \sqrt {\sin x} \leq \sqrt {\frac {\pi }{2}}, \] as desired.