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Let be $a$, $b$, $c$ sides of a triangle and $S$ his area. Prove that $$\sum_{\text{cyc}} \frac{1}{b^2+c^2+5bc-a^2} \leq \frac{\sqrt3}{8S}$$


My idea: $b^2+c^2-a^2 = 2bc \cos A$, so the inequality is equivalent to: $$\sum_{\text{cyc}} \frac{\sin A}{2\cos A+5}\leq\frac{\sqrt{3}}{4}$$

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  • $\begingroup$ From where does it come? $\endgroup$ – Dr. Sonnhard Graubner Sep 16 '18 at 8:12
  • $\begingroup$ i dont know..i have the problem from a friend..its for preparing olympiad $\endgroup$ – mathematiciangrade8 Sep 16 '18 at 8:21
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Now, you can use Jensen because $$\left(\frac{\sin\alpha}{5+2\cos\alpha}\right)''=\frac{\sin\alpha(10\cos\alpha-17)}{(5+2\cos\alpha)^3}<0$$ for all $0<\alpha<\pi.$

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  • $\begingroup$ thanks but i dont know what actually means '' $\endgroup$ – mathematiciangrade8 Sep 16 '18 at 8:26
  • $\begingroup$ (sin alpha / 5+2cos alpha ) ^ " ? $\endgroup$ – mathematiciangrade8 Sep 16 '18 at 8:26
  • $\begingroup$ I think he meant the second derivative with respect to $\alpha$ $\endgroup$ – Dr. Sonnhard Graubner Sep 16 '18 at 8:46
  • $\begingroup$ @mathematiciangrade8 It means that $f(x)=\frac{\sin{x}}{5+2\cos{x}}$ is a concave function and your inequality follows from the Jensen's inequality. $\endgroup$ – Michael Rozenberg Sep 16 '18 at 11:55

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