Prove that: $$\frac{a^{2}+b^{2}+c^{2}}{4\sqrt{3}S}+1 \ge 2\cdot\frac{4R+r}{\sqrt{3}p}$$
where $a, b, c$ are the sides of the triangle, $S$ is the area, $R$ is the circumradius, $r$ is the inradius and $p$ is the semi-perimeter.
My initial thoughts on the problem were comparing the numerators with the denominators and I have observed that $a^2 + b^2 + c^2 \geq 4\sqrt{3} S$, while $4R + r \geq p\sqrt{3}$, but these equations didn't lead to any reformulation, even though they both looked like useful known-lemmas.
$$\frac{a^{2}+b^{2}+c^{2}}{4\sqrt{3}S}+1 \ge 2\cdot\frac{4R+r}{\sqrt{3}p}$$ $$\frac{a^{2}+b^{2}+c^{2}}{4\sqrt{3}S}+1 \ge 2\cdot\frac{\frac{abc}{S}p+S}{\sqrt{3}p^2}$$ $$p^2(a^2+b^2+c^2)+4\sqrt3Sp^2\geq 8abcp+4S^2$$ Let's rephrase this using incircle tangent lengths, where $a=y+z,b=x+z,c=x+y,S^2=(x+y+z)xyz$ $$(x+y+z)(x^2+y^2+z^2+xy+yz+xz)+2\sqrt3(x+y+z)\sqrt{xyz(x+y+z)}\geq4(x+y)(y+z)(x+z)+2xyz$$ $$2\sqrt3(x+y+z)\sqrt{xyz(x+y+z)}\geq4(x+y)(y+z)(x+z)+2xyz-(x+y+z)(x^2+y^2+z^2+xy+yz+xz)$$ $$12(x+y+z)^2xyz(x+y+z)\geq(4(x+y)(y+z)(x+z)+2xyz-(x+y+z)(x^2+y^2+z^2+xy+yz+xz))^2$$ $$12(x+y+z)^2xyz(x+y+z) - (4(x+y)(y+z)(x+z)+2xyz-(x+y+z)(x^2+y^2+z^2+xy+yz+xz))^2\geq 0$$ We can use the symmetric polynomial theorem on this clearly symmetric polynomial to know it is a polynomial in $x+y+z,xy+yz+xz,xyz$.
I want to find the coefficient of $(xyz^2)$, so I'll substitute $1,\omega,\omega^2$. where $\omega^3=1$. we have $x+y+z=0,xy+yz+xz=0,xyz=1$. $$P(1,\omega,\omega^2)=-(...)^2$$ what's inside the parentheses is clearly real and thus the coefficient is negative.
Because of that, the minimal value is achieved when $xyz$ achieves extreme values while keeping $xy+yz+xz,x+y+z$ constant.
By uvw method, we have to check either when $x=0$, meaning the triangle is degenerate, $S=0$ and the inequality is clear or when $x=y$ which is when the triangle is isosceles which is a simple one variable inequality