<s>[INST] 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.
 [/INST] 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.
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