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.

Firstly using $s= \frac{1}{2} (a+b+c)$ the inequality is equivalent to $35(a^2 + b^2 + c^2) \geq 9(a+b+c)^2 + \frac{72abc}{a+b+c}$. Now note that $\frac{3abc}{a+b+c} = \frac{3}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}$ and so by the HM-GM-AM inequality it follows that $\frac{3}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}} \leq (abc)^\frac{2}{3} \leq \frac{1}{3}(a^2+b^2+c^2)$ hence we have $\frac{72abc}{a+b+c} \leq 8(a^2+b^2+c^2)$. Now using the AM-QM inequality $\frac{a+b+c}{3} \leq \sqrt{\frac{a^2 + b^2 + c^2}{3}} \Rightarrow 9(a+b+c)^2 \leq 27(a^2+b^2+c^2)$. Adding the two inequalities gives $35(a^2 + b^2 + c^2) \geq 9(a+b+c)^2 + \frac{72abc}{a+b+c}$ as required.
