<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] 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:
$$a^2+b^2+c^2+\frac{9abc}{2(a+b+c)}\geq\frac{(a+b+c)^2}{2}.$$

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