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.
 A: 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.
A: After expanding, the expression is equivalent to $$13(a^3+b^3+c^3)+4(a^2b+b^2c+c^2a)+4(ab^2+bc^2+ca^2)-63abc\geq 0$$
$$\Leftrightarrow 13[a^3+b^3+c^3-3abc]+4[(a^2b+b^2c+c^2a)-3abc]+4[(ab^2+bc^2+ca^2)-3abc]\geq 0,$$ which is clear, as follows from AM-GM.
[EDIT] With more thoughts and less work, it suffices by AM-GM to prove a stronger result: $$a^2+b^2+c^2\geq \frac{36}{35}\left(s^2+\frac{(2s/3)^3}s\right),\qquad (1)$$ where one replaces $abc$ by $((a+b+c)/3)^3=(2s/3)^3\geq abc.$
Clearly (1) is equivalent to $$3(a^2+b^2+c^2)\geq (a+b+c)^2,$$ which is true by C-S.
A: 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 $a^2+b^2+c^2 \geq ab+bc+ca$,
$13 (a^2+b^2+c^2) \geq 9(ab+bc+ca) + \cfrac{18abc}{s}$
$26 (a^2+b^2+c^2) \geq 18(ab+bc+ca) + \cfrac{36abc}{s}$
$35 (a^2+b^2+c^2) \geq 9(a^2+b^2+c^2) + 18(ab+bc+ca)  + \cfrac{36abc}{s}$
$35 (a^2 + b^2 + c^2) \geq 36 s^2 + \cfrac{36abc}{s}$
$a^2 + b^2 + c^2 \geq \cfrac{36}{35} \left(s^2 + \cfrac{abc}{s} \right)$
A: It's true for any positives $a$, $b$ and $c$.
Indeed, we need to prove that:
$$a^2+b^2+c^2\geq\frac{36}{35}\left(\frac{(a+b+c)^2}{4}+\frac{2abc}{a+b+c}\right)$$ or
$$35(a^2+b^2+c^2)(a+b+c)\geq9(a+b+c)^3+72abc$$ or
$$\sum_{cyc}(13a^3+4a^2b+4a^2c-21abc)\geq0,$$ which is true by Muirhead or by AM-GM.
The best result in this type it's the folowing.

Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$a^2+b^2+c^2+\frac{9abc}{2(a+b+c)}\geq\frac{(a+b+c)^2}{2}.$$

