inequation of sinuses of triangle's angles

Question

Let $\alpha, \beta, \gamma$ be inner angels of a triangle. Prove that the follwoing inequation is fulfilled: $\sin\alpha+\sin\beta+\sin\gamma-\sin\alpha\sin\beta\sin\gamma \ge \sin^3\alpha+\sin^3\beta+\sin^3\gamma$

Answer 1

we will prove that $$\sin(\alpha)+\sin(\beta)+\sin(\gamma)-\sin(\alpha)\sin(\beta)\sin(\gamma)\le \sin(\alpha)^3+\sin(\beta)^3+\sin(\gamma)^3$$ we use that $$\sin(\alpha)=\frac{a}{2R}$$ etc and $$A=\frac{abc}{4R}$$ and $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ plugging this in the given inequality we have to prove that $$a^3+b^3+c^3+abc-\frac{(abc)^2(a+b+c)}{4s(s-a)(s-b)(s-c)}\geq 0$$ with $$s=\frac{a+b+c}{2}$$ this is equivalent to $${\frac {{a}^{6}-{a}^{5}b-{a}^{5}c-{a}^{4}{b}^{2}+3\,{a}^{4}bc-{a}^{4}{ c}^{2}+2\,{a}^{3}{b}^{3}-2\,{a}^{3}{b}^{2}c-2\,{a}^{3}b{c}^{2}+2\,{a}^ {3}{c}^{3}-{a}^{2}{b}^{4}-2\,{a}^{2}{b}^{3}c+6\,{a}^{2}{b}^{2}{c}^{2}- 2\,{a}^{2}b{c}^{3}-{a}^{2}{c}^{4}-a{b}^{5}+3\,a{b}^{4}c-2\,a{b}^{3}{c} ^{2}-2\,a{b}^{2}{c}^{3}+3\,ab{c}^{4}-a{c}^{5}+{b}^{6}-{b}^{5}c-{b}^{4} {c}^{2}+2\,{b}^{3}{c}^{3}-{b}^{2}{c}^{4}-b{c}^{5}+{c}^{6}}{ \left( a+b -c \right) \left( a-b+c \right) \left( a-b-c \right) }} \le 0$$ now using the Ravi-substitution: $$a=y+z,b=x+z,c=x+y$$ and we get after simplification: $${\frac { \left( x-z \right) ^{2} \left( x+y \right) ^{2}}{-2\,z+x-y}}\le0$$ which is true.