<s>[INST] Ratio of sides of triangle $ABC$ If in a triangle $\Delta ABC$ with $a$, $b$ and $c$ as sides
$$\begin{align}\left(Cot\frac{A}{2}\right)^2 +\left(2Cot\frac{B}{2}\right)^2+\left(3Cot\frac{C}{2}\right)^2=\left(\frac{6s}{7r}\right)^2\end{align} \tag{1}$$ where $r$ is inradius and $s$ is SemiPerimeter, then find the ratio $a:b:c$
My try: If $\Delta$ is the area of the triangle we have
$$Cot\frac{A}{2}=\frac{s(s-a)}{\Delta}$$ and $$sr=\Delta$$
Using these in $(1)$ we get
$$ \frac{s^2(s-a)^2}{\Delta^2}+\frac{4s^2(s-b)^2}{\Delta^2}+\frac{9s^2(s-c)^2}{\Delta^2}=\frac{36s^4}{49\Delta^2}$$
$\implies$
$$49\left((s-a)^2+4(s-b)^2+9(s-c)^2\right)=36s^2$$
$\implies$
$$650s^2-98s(a+4b+9c)+49(a^2+4b^2+9c^2)=0$$
since for a triangle $s$ need to be unique, the above equation must have Discriminant Zero $\implies$
$$49(a+4b+9c)^2-650(a^2+4b^2+9c^2)=0$$
$\implies$
$$601a^2+1816b^2+1881c^2=392ab+3528bc+882ac$$
I am stuck up here, please help me. 
 [/INST] You have proved that the proposed condition is equivalent to
$$
(s-a)^2+4(s-b)^2+9(s-c)^2=\frac{36}{49}s^2\tag 1
$$
Note that, by the Cauchy-Schwarz inequality, we have
$$\eqalign{
s&=(s-a)+\frac{1}{2}(2(s-b))+\frac{1}{3}(3(s-c))\cr
&\leq\sqrt{(s-a)^2+4(s-b)^2+9(s-c)^2}\sqrt{1+\frac{1}{4}+\frac{1}{9}}\cr
&\leq\frac{7}{6}\sqrt{(s-a)^2+4(s-b)^2+9(s-c)^2}\cr
}$$
So $(1)$ corresponds to the case of equality in the Cauchy-Schwarz inequality, hence 
$$
\frac{s-a}{1}=\frac{2(s-b)}{\frac{1}{2}}=\frac{3(s-c)}{\frac{1}{3}}
$$
that is
$$
s-a=\lambda,\quad s-b=\frac{\lambda}{4},\quad s-c=\frac{\lambda}{9}
$$
for some $\lambda>0$. Adding these equalities we get: $s=\frac{49}{36}\lambda$,
that is
$$
a=\frac{13}{36}\lambda,\quad b=\frac{40}{36}\lambda,\quad c=\frac{45}{36}\lambda
$$
Finally $a:b:c=13:40:45$.$\qquad\square$
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