Problems

A function \(\f(x)\) is said to be 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 \(a< x < b\,\).

1. 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 \(a< w < b\,\).
2. Show that, if \(A\), \(B\) and \(C\) are the angles of a triangle, then \[ \sin A +\sin B + \sin C \le \frac{3\sqrt3}2 \,. \]
3. By considering \(\ln (\sin x)\), show that, if \(A\), \(B\) and \(C\) are the angles of a triangle, then \[ \sin A \times \sin B \times \sin C \le \frac {3 \sqrt 3} 8 \,. \]

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Solution:

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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.

1. \(\,\) \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*}
2. 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 \(\sin A + \sin B + \sin C \leq 3 \sin \frac{A+B+C}{3} = 3 \sin \frac{\pi}{3} = \frac{3\sqrt{3}}2\)
3. 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*}