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(Angle)^(Side)[Inequality]
Posted: Sun Oct 19, 2014 10:42 pm
by sowmitra
Let, $ABC$ be a triangle with angles $A,B,C$, and, sides $a,b,c$ (usual notations). Let, $R$ be the circum-radius. Prove that, \[\left(\frac{2A}{\pi}\right)^\frac{1}{a}\left(\frac{2B}{\pi}\right)^\frac{1}{b}\left(\frac{2C}{\pi}\right)^\frac{1}{c}\leq \left(\frac{2}{3}\right)^{\frac{\sqrt{3}}{R}}\]
Re: (Angle)^(Side)[Inequality]
Posted: Mon Oct 20, 2014 9:08 am
by *Mahi*
$\log$ - Chebyshev - Jensen - Sin law - Jensen - Exponent
Re: (Angle)^(Side)[Inequality]
Posted: Mon Oct 20, 2014 8:26 pm
by sowmitra
I think one Jensen can suffice...
Re: (Angle)^(Side)[Inequality]
Posted: Mon Oct 20, 2014 11:16 pm
by *Mahi*
On which function?
Re: (Angle)^(Side)[Inequality]
Posted: Tue Oct 21, 2014 12:04 am
by sowmitra
\[f(x)=\frac{1}{\sin{x}}\cdot\ln\left(\frac{2x}{\pi}\right)\]
Re: (Angle)^(Side)[Inequality]
Posted: Tue Oct 21, 2014 10:16 am
by *Mahi*
This function is concave for $x \in \left (0, \frac \pi 2 \right )$ and convex for $x \in \left ( \frac \pi 2 , \pi \right )$.
Re: (Angle)^(Side)[Inequality]
Posted: Tue Oct 21, 2014 11:19 pm
by Nirjhor
Re: (Angle)^(Side)[Inequality]
Posted: Wed Oct 22, 2014 12:19 am
by sowmitra
Oops... sorry
I miscalculated by taking $\frac{1}{x}$ instead of $\frac{1}{\sin x}$ while differentiating...