\(% 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}}\)

2018.2.2 Question 2

If \(f''(x) < 0\), this means \(f'(x)\) is decreasing, i.e. the gradient of a tangent to the curve \(y = f(x)\) is decreasing. Assume, B.W.O.C., that some \(f(x)\) satisfies this condition but is not convex. This means that there exists some \(a < x_1 < x_2 < b\) and some \(0 < t < 1\) that \[ t f(x_1) + (1 - t) f(x_2) \geq f(tx_1 + (1 - t) x_2). \]

This means that some point on the chord connecting \((x_1, f(x_1))\) and \((x_2, f(x_2))\) is above the graph of the function at that point with \(x\)-coordinate \(t x_1 + (1 - t) x_2\). Hence, the gradient of that function must be less than the gradient of the chord at that point, and since \(f''(x) < 0\), the function must continue to have a gradient of less than this, and hence cannot pass through \((x_2, f(x_2))\).

Hence, this triple of \((x_1, x_2, t)\) does not exist, and the function \(f\) must be concave on \((a, b)\).

1. Let \(x_1 = \frac {2u + v}{3}\) and \(x_2 = \frac {v + 2w}{3}\), and let \(t = \frac {1}{2}\). We can see that \(a < x_1, x_2 < b\) and hence we have \[ \frac {1}{2} f(x_1) + \frac {1}{2} f(x_2) \leq f\left (\frac {1}{2} x_1 + \frac {1}{2} x_2\right ), \] which gives \[ \frac {1}{2} f\left (\frac {2u + v}{3}\right ) + \frac {1}{2} f\left (\frac {v + 2w}{3}\right ) \leq f\left (\frac {u + v + w}{3}\right ). \]

   Let \(x_1 = u\) and \(x_2 = v\), and let \(t = \frac {2}{3}\). We have \[ \frac {2}{3} f(u) + \frac {1}{3} f(v) \leq f\left (\frac {2u + v}{3}\right ), \] and let \(x_1 = w\), \(x_2 = v\), and let \(t = \frac {2}{3}\), we have \[ \frac {2}{3} f(w) + \frac {1}{3} f(v) \leq f\left (\frac {2w + v}{3}\right ). \]

   Hence, \begin {align*} f\left (\frac {u + v + w}{3}\right ) & \geq \frac {1}{2} f\left (\frac {2u + v}{3}\right ) + \frac {1}{2} f\left (\frac {v + 2w}{3}\right ) \\ & \geq \frac {1}{2} \cdot \left [\frac {2}{3} f(u) + \frac {1}{3} f(v)\right ] + \frac {1}{2} \cdot \left [\frac {2}{3} f(w) + \frac {1}{3} f(v)\right ] \\ & = \frac {1}{3} \left [f(u) + f(v) + f(w)\right ], \end {align*}

   which shows exactly what is desired.

2. Let \(a = 0\) and \(b = \pi \), and let \(f(x) = \sin x\). We aim to show that \(f\) is concave, and notice that \[ f''(x) = - \sin x < 0 \] for all \(0 < x < \pi \), so it is concave on \((0, \pi )\).

   Angles in a triangle lie within \((0, \pi )\), and they must sum up to \(\pi \). Hence, by applying the previous part, we have \[ \sin A + \sin B + \sin C \leq 3\sin \left (\frac {A + B + C}{3}\right ) = 3\sin \left (\frac {\pi }{3}\right ) = \frac {3 \sqrt {3}}{2}, \] as desired.

3. We keep \(a = 0\) and \(b = \pi \), and let \(f(x) = \ln \sin x\). Note that \[ f'(x) = \frac {\cos x}{\sin x} = \cot x, \] and hence \[ f''(x) = - \csc ^2 x < 0 \] which shows that \(f\) is concave on \((0, \pi )\).

   Hence, \begin {align*} \ln (\sin A \sin B \sin C) & = \ln \sin A + \ln \sin B + \ln \sin C \\ & \leq 3 \ln \sin \left (\frac {A + B + C}{3}\right ) \\ & = 3 \ln \sin \left (\frac {\pi }{3}\right ) \\ & = 3 \ln \frac {\sqrt {3}}{2} \\ & = \ln \frac {3 \sqrt {3}}{8}. \end {align*}

   Since \(\ln \) is a strictly increasing function, we can then conclude that \[ \sin A \sin B \sin C \leq \frac {3\sqrt {3}}{8}, \] as desired.