Question

IF ABC IS an acute triangle show that:

Answer 1

\[\sin \frac{ A }{ 2 }\sin \frac{ B }{ 2 }\sin \frac{ C }{ 2 }\le \frac{ 1 }{ 8 }\]

Answer 2

@eliassaab @thomaster @undeadknight26 @ganeshie8 @

Answer 3

Here is a proof with Mathematica

Answer 4

The maximum 1/8 is attained when
A=B=C=60 degrees

Answer 5

The way to do it properly is the following. Notice that
\[
f(x) = \ln(\sin(x))
\]
is concave on the interval \([0, \pi] \)

Answer 6

Hence
\[
\frac 1 3 f(A/2) +\frac 1 3 f(B/2)+ \frac 1 3 f(C/2)\le
f\left( \frac 1 3 \frac A 2 +\frac 1 3 \frac B 2 +\frac 1 3 \frac C 2 \right)=\\

f\left( \frac 1 3 \left( \frac A 2 +\frac B 2 +\frac C 2 \right) \right)=\\
f\left( \frac 1 3 \left( \frac \pi 2 \right) \right)=f\left( \frac \pi 6 \right)=

\ln(\sin(\pi/6)=\ln(1/2)\\
\]
The rest is easy

Answer 7

\[
\frac 1 3 f(A/2) +\frac 1 3 f(B/2)+ \frac 1 3 f(C/2)=\\

\frac 1 3 \ln( \sin((A/2)) +\frac 1 3 \ln(\sin(B/2))+ \frac 1 3 \ln\sin((C/2))=\\

\frac 1 3 \left(\ln( \sin((A/2)) + \ln(\sin(B/2))+ \ln\sin((C/2))\right)=\\
\frac 1 3 \left(\ln( \sin((A/2) \sin((B/2)\sin((C/2)) \right)=\\
\ln( \sin((A/2) \sin((B/2)\sin((C/2))^{1/3}\le \ln(1/2)\\
\sin((A/2) \sin((B/2)\sin((C/2))^{1/3}\le \frac 12 \\
\sin((A/2) \sin((B/2)\sin((C/2))\le \left(\frac 12\right)^3=\frac 1 8 \\
\]

Answer 8

@ganeshie8

Answer 9

Actually this exercise is strictly related to jessner theorem

Answer 10

beautiful !

another way using lagrange multipliers :

\(\large f : A+ B + C = \pi \)
\(\large g : \sin \frac{A}{2} \sin \frac{B}{2}\sin \frac{C}{2}\)

\(\large \nabla g = \lambda \nabla f\)

solving gives the max value of \(\large g= \frac{1}{8}\) at \(\large A = B = C = \frac{\pi}{3}\)