In $\Delta$$ABC$, $a$ $\ge$$b$$\ge$$c$. If $$\frac{a^3+b^3+c^3}{sin^3 A+sin^3 B+ sin^3 C}=7 $$ compute the maximum value of $a$.
My efforts so far:
By law of sines i was able to get the following identities:
$a^3=7sin^3A$ , $b^3=7sin^3B$ and $c^3=7sin^3C$
But i don't know how to manipulate these identities in some kind of usefull way, i think i am proceeding in the wrong track. The problem clearly asks for some tricky manipulation of the equation above,such that applying the restrictions for $a,b,c$ i can not have $a$ greater than some defined value,but ,as i already said, i don't see how to do that.
Important: I would really appreciate if you could give me some hints and not the solution right away(maybe you can spoiler it if you want with hints above), otherwise my efforts would've been vains.
That said,thanks in advance.
*(Editing solution in case someone else in the future will need)
Solution:
1)Given $a^3=7sin^3A$ ,you have a maximum when $\sin A =1$,hence $a=7^{1/3}$.
2)This problem can also be solved in the following way: $$\frac{a^3+b^3+c^3}{sin^3 A+ sin^3 b +sin^3 C} = \frac{a^3+b^3+c^3}{\frac{a^3+b^3+c^3}{8R^3}}$$
Solve for R,and then you equate $a=2R$ (a is equal to the diameter of the circumcircle)
It seems that you have to make $\sin A$ as large as possible.