\(% 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}}\)
We notice that \begin {align*} \DiffFrac {E}{x} & = 2 \cdot \DiffFrac {y}{x} \cdot \NdiffFrac {2}{y}{x} + 2 y^3 \DiffFrac {y}{x} \\ & = 2 \cdot \DiffFrac {y}{x} \cdot \left (\NdiffFrac {2}{y}{x} + y^3\right ) \\ & = 0, \end {align*}
and so \(E\) must be constant.
So hence \begin {align*} E(x) & = E(0) \\ & = 0^2 + \frac {1}{2} \\ & = \frac {1}{2}. \end {align*}
Therefore, \[ y^4 = 2\left [E(x) - \left (\DiffFrac {y}{x}\right )^2\right ] \leq 2 E(x) = 1, \] and hence \[ \abs *{y(x)} \leq 1. \]
We notice that \begin {align*} \DiffFrac {E}{x} & = 2 \cdot \DiffFrac {v}{x} \cdot \NdiffFrac {2}{v}{x} + 2 \sinh v \DiffFrac {v}{x} \\ & = 2 \DiffFrac {v}{x} \cdot \left (\NdiffFrac {2}{v}{x} + \sinh v\right ) \\ & = 2 \DiffFrac {v}{x} \cdot \left (-x \DiffFrac {v}{x}\right ) \\ & = -2x \left (\DiffFrac {v}{x}\right )^2, \end {align*}
so when \(x \geq 0\), since \(\left (\DiffFrac {v}{x}\right )^2 \geq 0\), we must have \[ \DiffFrac {E}{x} \leq 0. \]
Therefore, for \(x \geq 0\), \(E(x) \leq E(0) = 0^2 + 2 \cosh \ln 3 = 3 + \frac {1}{3} = \frac {10}{3}\). Hence, \begin {align*} \cosh v(x) & = \frac {E(x) - \left (\DiffFrac {v}{x}\right )^2}{2} \\ & \leq \frac {\frac {10}{3}}{2} \\ & = \frac {5}{3}. \end {align*}
Notice that \begin {align*} \DiffOp {x} \left (\DiffFrac {w}{x}\right )^2 & = 2 \cdot \DiffFrac {v}{x} \cdot \NdiffFrac {2}{v}{x} \\ & = - 2 \cdot \DiffFrac {w}{x} \cdot \left [\left (5 \cosh x - 4 \sinh x - 3\right ) \cdot \DiffFrac {w}{x} + \left (w \cosh w + 2 \sinh w\right )\right ]. \end {align*}
We also notice that \begin {align*} \int (w \cosh w + 2 \sinh w) \Diff w & = \int w \cosh w \Diff w + 2 \cosh w \\ & = \int w \Diff \sinh w + 2 \cosh w + C \\ & = w \sinh w - \int \sinh w \Diff w + 2 \cosh w + C \\ & = w \sinh w - \cosh w + 2 \cosh w + C \\ & = w \sinh w + \cosh w + C, \end {align*}
so consider the function \[ E(x) = \left (\DiffFrac {w}{x}\right )^2 + 2(w \sinh w + \cosh w), \] and we have \begin {align*} \DiffFrac {E}{x} & = - 2 \cdot \DiffFrac {w}{x} \cdot \left [\left (5 \cosh x - 4 \sinh x - 3\right ) \cdot \DiffFrac {w}{x} + \left (w \cosh w + 2 \sinh w\right ) - \left (w \cosh w + 2 \sinh w\right )\right ] \\ & = -2 \left (\DiffFrac {w}{x}\right )^2 \left (5 \cosh x - 4 \sinh x - 3\right ) \\ & = - \left (\DiffFrac {w}{x}\right )^2 \left [5 \left (e^x + e^{-x}\right ) - 4 \left (e^x - e^{-x}\right ) - 6\right ] \\ & = - \left (\DiffFrac {w}{x}\right )^2 \left (e^x + 9e^{-x} - 6\right ) \\ & = - e^{-x} \left (\DiffFrac {w}{x}\right )^2 (e^x - 3)^2 \\ & \leq 0. \end {align*}
Hence, \[ E(x) \leq E(0) = \left (\frac {1}{\sqrt {2}}\right )^2 + 2 (0 \sinh 0 + \cosh 0) = \frac {1}{2} + 2 = \frac {5}{2}, \] for \(x \geq 0\).
Therefore, \[ \frac {5}{2} \geq \left (\DiffFrac {w}{x}\right )^2 + 2(w \sinh w + \cosh w), \] and hence \[ 2(w \sinh w + \cosh w) \leq \frac {5}{2} \] for \(x \geq 0\) since squares are always non-negative.
Hence, \[ \cosh w \leq \frac {5}{4} - w \sinh w \leq \frac {5}{4} \] for \(x \geq 0\), the second inequality being true since \(w \sinh w \geq 0\) since \(\sinh w\) and \(w\) always take the same sign, as desired.