# Show that $\sum_{\text{cyc}} \frac{1}{b^2+c^2+5bc-a^2} \leq \frac{\sqrt3}{8S}$ for a triangle with sides $a$, $b$, $c$ and area $S$

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: $$\sum_{\text{cyc}} \frac{\sin A}{2\cos A+5}\leq\frac{\sqrt{3}}{4}$$

• From where does it come? – Dr. Sonnhard Graubner Sep 16 '18 at 8:12
• i dont know..i have the problem from a friend..its for preparing olympiad – mathematiciangrade8 Sep 16 '18 at 8:21

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 $0<\alpha<\pi.$
• I think he meant the second derivative with respect to $\alpha$ – Dr. Sonnhard Graubner Sep 16 '18 at 8:46
• @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. – Michael Rozenberg Sep 16 '18 at 11:55