(Angle)^(Side)[Inequality]
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]
$\log$ - Chebyshev - Jensen - Sin law - Jensen - Exponent ![Smile :)](./images/smilies/icon_e_smile.gif)
![Smile :)](./images/smilies/icon_e_smile.gif)
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Re: (Angle)^(Side)[Inequality]
I think one Jensen can suffice... ![Wink ;)](./images/smilies/icon_e_wink.gif)
![Wink ;)](./images/smilies/icon_e_wink.gif)
Re: (Angle)^(Side)[Inequality]
On which function?
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Re: (Angle)^(Side)[Inequality]
\[f(x)=\frac{1}{\sin{x}}\cdot\ln\left(\frac{2x}{\pi}\right)\]
Re: (Angle)^(Side)[Inequality]
This function is concave for $x \in \left (0, \frac \pi 2 \right )$ and convex for $x \in \left ( \frac \pi 2 , \pi \right )$.
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Re: (Angle)^(Side)[Inequality]
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
Re: (Angle)^(Side)[Inequality]
Oops... sorry
I miscalculated by taking $\frac{1}{x}$ instead of $\frac{1}{\sin x}$ while differentiating... ![Confused :?](./images/smilies/icon_e_confused.gif)
![Embarrassed :oops:](./images/smilies/icon_redface.gif)
![Confused :?](./images/smilies/icon_e_confused.gif)