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Problem Of The Week #412 Apr 12th, 2020

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Here is this week's POTW:

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Prove that in a triangle with sides $a, b$ and $c$ and opposite angles $A, B$ and $C$ (in radians), the following relation holds:

$\dfrac{aA+bB+cC}{a+b+c}\ge\dfrac{\pi}{3}$

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Congratulations to lfdahl for his correct solution(Cool), which you can find below:

The result is a direct consequence of Chebyschev´s sum inequality:

For any triangle we can WLOG choose: $a \geq b\geq c$ - which implies: $A \geq B \geq C$ (radians with total angle sum $\pi$).

Applying CSI
\[\frac{1}{3}\left ( aA+bB+cC \right ) \geq \frac{1}{3}\left (a+b+c \right ) \frac{1}{3}\left ( A+B+C \right )\]

or

\[\frac{aA+bB+cC }{a+b+c} \geq \frac{\pi}{3}\]
 
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