Triangle and Maxium value

Question

Given any triangle ABC with $a \ge b \ge c$ such that $\frac{a^3+b^3+c^3}{\sin^3(A)+\sin^3(B)+\sin^3(C)}=7$, what is the maximum value of $a$?

Answer 1

since $$\sin^3{A}+\sin^3{B}+\sin^3{C}=(8R^3)^{-1}(a^3+b^3+c^3)$$ so $$(8R^3)^{-}=\dfrac{1}{7}\Longrightarrow R^{-1}=\sqrt[3]{\dfrac{1}{56}}$$ so $$a=2R\sin{A}\le 2R=\sqrt[3]{7}$$

Answer 2

Using the Sine Law, i.e. $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k $$

we find that the condition in the question is merely equivalent to $k^3 = 7$.

So $a = k \sin A \le k = \sqrt[3]7$.