Problem source

A  function $\f(x)$ is said to be \textit{concave} for $a<   x <   b$
if
\[
 \ t\,\f(x_1) +(1-t)\,\f(x_2)
\le 
\f\big(tx_1+ (1-t)x_2\big) 
\,
,\]   for $a< x_1 < b\,$, $a<   x_2<   b$ and $0\le t \le 1\,$.
Illustrate this definition by means of a sketch, showing the chord joining the points $\big(x_1, \f(x_1)\big) $ and $\big(x_2, \f(x_2)\big) $, in the case $x_1 < x_2$ and $\f(x_1)< \f(x_2)\,$.
Explain why a function $\f(x)$ satisfying  $\f''(x)<0$ for $a<   x <   b$  is concave for $a<   x <   b\,$.
\begin{questionparts}
\item
 By choosing $t$, $x_1$ and $x_2$ suitably, show that, if $\f(x)$ is 
concave for $a<   x <   b\,$, then
\[
\f\Big(\frac{u+ v+w}3\Big) \ge \frac{ \f(u) +\f(v) +\f(w)}3
\,
,\]  
for  $a<   u <   b\,$,  $a<   v <   b\,$ and  $a<   w <   b\,$.
\item
Show that,  if $A$, $B$ and $C$ are the  angles of a triangle, then
\[
\sin A +\sin B + \sin C \le \frac{3\sqrt3}2
\,.
\]
\item By considering $\ln (\sin x)$, show that, if $A$, $B$ and $C$ are the  angles of a triangle, then
\[
\sin A \times \sin B \times \sin C \le \frac {3 \sqrt 3}  8 \,.
\]
\end{questionparts}

Solution source

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){(#1)^2-(#1)+1};
    \def\xl{-1};
    \def\xu{4};
    \def\yl{-1};
    \def\yu{10};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);

        \filldraw (0.5, {\functionf(0.5)}) circle (1pt) node[below] {$(x_1, f(x_1))$};
        \filldraw (2.5, {\functionf(2.5)}) circle (1pt) node[above] {$(x_2, f(x_2))$};

        \filldraw (0.4 * 0.5 + 0.6 * 2.5, 0) circle (1.5pt) node[below] {$tx_1+(1-t)x_2$};
        \filldraw (0.4 * 0.5 + 0.6 * 2.5, {0.4 * (\functionf(0.5)) + 0.6 * (\functionf(2.5))}) circle (1pt) node[above, rotate=45] {\small $tf(x_1)+(1-t)f(x_2)$};
        \filldraw (0.4 * 0.5 + 0.6 * 2.5, {\functionf(0.4 * 0.5 + 0.6 * 2.5)}) circle (1pt) node[below, rotate=45] {\small $ f(tx_1+(1-t)x_2)$};

        \draw[thick] (0.5, {\functionf(0.5)}) -- (2.5, {\functionf(2.5)});
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {\functionf(\x)});

    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}


Consider the function $g(t) = f(tx_1 + (1-t)x_2) - tf(x_1) - (1-t)f(x_2)$, notice that $g(0) = g(1) = 0$. Since $g''(x) < 0$ over the whole interval, we must have two things:

1. $g'(x)$ is increasing.
2. It $g'(x) = 0$ can have at most one solution.

Therefore $g'(x)$ is initially $0$, we have exactly one turning point. Therefore the function is initially decreasing and then increasing, therefore it is always negative and our inequality holds.

\begin{questionparts}

\item $\,$
\begin{align*}
&& f \left ( \frac{u+v+w}{3} \right) &= f \left ( \frac{2}{3}\cdot \frac{u+v}2+\frac{1}{3}w \right) \\
&&&\geq \frac23 f \left ( \frac{u+v}{2} \right) + \frac13 f(w) \\
&&&\geq \frac23 \left (\frac12 f(u) + \frac12 f(v) \right) + \frac13 f(w) \\
&&&= \frac{f(u)+f(v)+f(w)}{3}
\end{align*}

\item Notice that if $A, B, C$ are angles in a triangle then they add to $\pi$ $0 < A,B,C < \pi$. We also have $f(x) = \sin x \Rightarrow f''(x) = - \sin x < 0$ on this interval. Therefore $\sin A + \sin B + \sin C \leq 3 \sin \frac{A+B+C}{3} = 3 \sin \frac{\pi}{3} = \frac{3\sqrt{3}}2$

\item Also notice that 
\begin{align*}
&& f(x) &= \ln ( \sin x) \\
\Rightarrow && f'(x) &= \frac{\cos x}{ \sin x} \\
&& f''(x) &= -\textrm{cosec}^2 x < 0 \\
\\
\Rightarrow && \ln( \sin A) + \ln (\sin B) + \ln (\sin C) &\leq 3 \ln \left (\sin \left ( \frac{A + B+ C}{3} \right) \right) \\
&&&= 3 \ln \left ( \frac{\sqrt{3}}{2} \right) = \ln \frac{3\sqrt{3}}{8} \\
\Rightarrow && \sin A \sin B \sin C &\leq \frac{3\sqrt{3}}8
\end{align*}
\end{questionparts}