Proving a triangle equilateral given condition $al_a^2+bl_b^2+cl_c^2=9R\Delta$ $ABC$ is a triangle, with $l_a$, $l_b$, $l_c$ as angle bisectors, $R$ as circumradius and $\Delta$ as area, such that:
$$al_a^2+bl_b^2+cl_c^2=9R\Delta$$ 
Is it true that $ABC$ is equilateral?
I am not sure how to approach this problem. Ideas, anyone? :)
 A: Use the fact that $l_a$ divides the side $BC = a$ (at $A'$) in the ratio $BA':CA' = b:c$, and from Stewart's Theorem we have $\displaystyle l_a^2 = bc - \frac{a^2bc}{(b+c)^2}$. 
Thus, $\displaystyle \sum\limits_{cyc} al_a^2 = 3abc - abc\sum\limits_{cyc}\frac{a^2}{(b+c)^2} = 9R\Delta = \frac{9}{4}abc \implies \sum\limits_{cyc}\frac{a^2}{(b+c)^2} = \frac{3}{4}$
Now by Cauchy-Schwarz: $\displaystyle \sum\limits_{cyc}\frac{a^2}{(b+c)^2} \ge \frac{1}{3}\left(\sum\limits_{cyc}\frac{a}{b+c}\right)^2$ and $\displaystyle R.H.S. \ge \frac{3}{4}$ from Nesbitt's Inequality.
Equality holds iff $a=b=c$. Hence, the triangle must be equilateral.
A: Write everything in terms of $a,b,c$.
A few key formulas,
$R=\frac{abc}{4\Delta}$, this comes from $\frac{a}{\sin(A)}=2R$ and $\Delta=\frac{1}{2}bc\sin(A)$.
$l_a^2=c^2+(\frac{ab}{b+c})^2-2c(\frac{ab}{b+c})\cos(B)$, this is a combination of law of cosine and angle bisector theorem. Use law of cosine again to expand $\cos(B)$. Write similar formulas for $l_b$ and $l_c$, the rest of the question is just checking the algebra.
