(Angle)^(Side)[Inequality]

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sowmitra
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(Angle)^(Side)[Inequality]

Unread post by sowmitra » Sun Oct 19, 2014 10:42 pm

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, \[\left(\frac{2A}{\pi}\right)^\frac{1}{a}\left(\frac{2B}{\pi}\right)^\frac{1}{b}\left(\frac{2C}{\pi}\right)^\frac{1}{c}\leq \left(\frac{2}{3}\right)^{\frac{\sqrt{3}}{R}}\]
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Re: (Angle)^(Side)[Inequality]

Unread post by *Mahi* » Mon Oct 20, 2014 9:08 am

$\log$ - Chebyshev - Jensen - Sin law - Jensen - Exponent :)
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sowmitra
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Re: (Angle)^(Side)[Inequality]

Unread post by sowmitra » Mon Oct 20, 2014 8:26 pm

I think one Jensen can suffice... ;)
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Re: (Angle)^(Side)[Inequality]

Unread post by *Mahi* » Mon Oct 20, 2014 11:16 pm

On which function?
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sowmitra
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Re: (Angle)^(Side)[Inequality]

Unread post by sowmitra » Tue Oct 21, 2014 12:04 am

\[f(x)=\frac{1}{\sin{x}}\cdot\ln\left(\frac{2x}{\pi}\right)\]
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Re: (Angle)^(Side)[Inequality]

Unread post by *Mahi* » Tue Oct 21, 2014 10:16 am

This function is concave for $x \in \left (0, \frac \pi 2 \right )$ and convex for $x \in \left ( \frac \pi 2 , \pi \right )$.
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Re: (Angle)^(Side)[Inequality]

Unread post by Nirjhor » Tue Oct 21, 2014 11:19 pm

Inverting and taking $\ln$ gives us to prove \[\dfrac{1}{a} \ln\dfrac{\pi}{2A}+\dfrac{1}{b} \ln\dfrac{\pi}{2B}+\dfrac{1}{c} \ln\dfrac{\pi}{2C}\ge \dfrac{\sqrt 3}{R}\ln \dfrac{3}{2}.\] Jensen's on convex $f(x)=\ln\dfrac{\pi}{2x}$ gives \[\ln\dfrac{\pi}{2A}+\ln\dfrac{\pi}{2B}+\ln\dfrac{\pi}{2C}\ge 3\ln\left(\dfrac{3\pi}{2A+2B+2C}\right)=3\ln\dfrac{3}{2}.\] Now Jensen's on concave \(y=\sin x\) on $[0,\pi]$ gives \[a+b+c=2R\left(\sin A+\sin B+\sin C\right)\le 6R\sin\dfrac{\pi}{3}=3R\sqrt 3.\] So by Titu's lemma \[\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge \dfrac{3^2}{a+b+c}=\dfrac{9}{a+b+c}\ge \dfrac{\sqrt 3}{R}.\] Finally by Chebyshev's and applying the above result \[\begin{eqnarray} \dfrac{1}{a} \ln\dfrac{\pi}{2A}+\dfrac{1}{b} \ln\dfrac{\pi}{2B}+\dfrac{1}{c} \ln\dfrac{\pi}{2C} &\ge & \dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\ln\dfrac{\pi}{2A}+\ln\dfrac{\pi}{2B}+\ln\dfrac{\pi}{2C}\right) \\ \\ &\ge & \left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ln\dfrac{3}{2} ~\ge \dfrac{\sqrt 3}{R} \ln\dfrac{3}{2}\end{eqnarray}\] which is what we need to save the world.
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sowmitra
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Re: (Angle)^(Side)[Inequality]

Unread post by sowmitra » Wed Oct 22, 2014 12:19 am

Oops... sorry :oops: I miscalculated by taking $\frac{1}{x}$ instead of $\frac{1}{\sin x}$ while differentiating... :?
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Some-Angle Related Problems;

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