A geometric inequality to triangle $ABC$

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If $a, b, c$ be the sides of a triangle $ABC$, I have to prove that

$\sum\limits_{cycl}^{} (\frac {a}{b+c})^2 + \frac {3 A^2}{abcs} \ge \frac {9}{8}$

where $A,s$ the area and the semiperemeter. I tried to use the Ravi's substitution, but the inequality was then quite hard to solve. It might be a differente method. Thank you