Prove that if $a,b,c$ are sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$ 
Prove that, if $a,b,c$ are the sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$.

What I Tried:- Nothing special really came in my mind. I did not find a way to use Triangle Inequality. What I did was, by AM-GM :-
$$a^2 + b^2 + c^2 \geq \frac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg) > \frac{72}{35}(\sqrt{sabc}).$$
But I couldn't proceed from this.
Another Idea I had was :- $a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)$. 
$\rightarrow a^2 + b^2 + c^2 = 4s^2 - 2(ab + bc + ca).$
But I did not know how to use this here, and would make the calculations a bit messy, especially of the $\dfrac{36}{35}$ part present there.
Can Anyone Help me? Thank You.

Here is another approach.
By AM-GM,
$(a+b+c)(a^2+b^2+c^2) \geq 3 (abc)^{2/3} \cdot 3 (abc)^{1/3} = 9abc$
$4 (a^2+b^2+c^2) \geq \cfrac{18abc}{s}$
$13 (a^2+b^2+c^2) \geq 9(a^2+b^2+c^2) + \cfrac{18abc}{s}$
Using $a^2+b^2+c^2 \geq ab+bc+ca$,
$13 (a^2+b^2+c^2) \geq 9(ab+bc+ca) + \cfrac{18abc}{s}$
$26 (a^2+b^2+c^2) \geq 18(ab+bc+ca) + \cfrac{36abc}{s}$
$35 (a^2+b^2+c^2) \geq 9(a^2+b^2+c^2) + 18(ab+bc+ca)  + \cfrac{36abc}{s}$
$35 (a^2 + b^2 + c^2) \geq 36 s^2 + \cfrac{36abc}{s}$
$a^2 + b^2 + c^2 \geq \cfrac{36}{35} \left(s^2 + \cfrac{abc}{s} \right)$
