help; trigonometric inequality

J.Y.Choi

If $A,B,C$ are three angles of a triangle, prove that :

${\sin{A}}+{\sin{B}}+{\sin{C}}-4{\sin{A}}{\sin{B}}{\sin{C}}\geq{0}$ .

One of my trying was thinking of
$\sin{3A}+\sin{3B}+\sin{3C}=3\left(\sin{A}+\sin{3B}+\sin{3C}-4\cdot\frac{\sin^3{A}+\sin^3{B}+\sin^3{C}}{3}\right)\leq3\left(\s...$ .
But it's wrong that $\quad\sin{3A}+\sin{3B}+\sin{3C}\geq{0}$ .....Please, help me. Thanks.

**Posted:** Fri Nov 06, 2009 9:25 am

Dr Sonnhard Graubner

hello, using that
$\sin(A)+\sin(B)+\sin(C)=4\cos(A/2)\cos(B/2)\cos(C/2)$ your inequality is equivalent to
$\sin(A/2)\sin(B/2)\sin(C/2)\le\frac{1}{8}$
and this is equivalent to
$abc\geq(-a+b+c)(a-b+c)(a+b-c)$ which is well-known.
Sonnhard.

**Posted:** Fri Nov 06, 2009 12:14 pm

J.Y.Choi

Thanks again for your nice solution.

I just had another solution(not mine) :

Let $a,b,c$ be three sides and $S$ be the area of $\triangle{ABC}$ respectively. By law of sines, it is equivalent to

$\frac{a}{2R}+\frac{b}{2R}+\frac{c}{2R}-4\cdot\frac{a}{2R}\cdot\frac{b}{2R}\cdot\frac{c}{2R}\geq{0}$ . Product $2R$ to each side and by $S=\frac{abc}{4R}=\frac{1}{2}r(a+b+c)$ , it is

$\frac{2S}{r}-\frac{4}{R}S\geq{0}$ . Where $r$ is the radius of inscribed circle of $\triangle{ABC}$ .
Now, we can see the inequality we want to prove is equivalent to Euler's triangle inequality $R\geq{2r}$ .

**Posted:** Sat Nov 07, 2009 7:00 pm