In the triangle $ABC$, $a$, $b$ and $c$ are the sides opposite to vertices $A$, $B$, $C$.
$a$, $b$ and $c$ are natural Numbers and their $HCF$ is $1$.
If $$\dfrac{\left ( s-a \right )^{2}}{36}+\dfrac{\left ( s-b \right )^{2}}{9}+\dfrac{\left ( s-c \right )^{2}}{4}=\dfrac{s^{2}}{49}$$ where $s$ is the semi_perimeter, then find the side lengths.
Now here is my take. After rewriting the given expression $$ 49\left \{ \left ( s-a \right )^{2}+4\left ( s-b \right )^{2}+9\left ( s-c \right )^{2} \right \}=36s^{2} $$ we can see here $s^{2}$ is divisible by $49$.
I can not go further perhaps my number theory is very bad. Please help me.
Using Cauchy-Schwarz:
$$\dfrac{\left ( s-a \right )^{2}}{36}+\dfrac{\left ( s-b \right )^{2}}{9}+\dfrac{\left ( s-c \right )^{2}}{4}\geq \frac{(3s-a-b-c)^2}{36+9+4}=\dfrac{s^{2}}{49}$$
Equality occurs if and only if
$$\frac{s-a}{36}=\frac{s-b}{9}=\frac{s-c}{4}$$
Can you end it from here?