We have $a^2\geq 2a+3-36\left(\frac{1}{a}-\frac{1}{3}\right)\iff(a-3)^2(a+4)\geq0$ Summing the similar inequalities for $a,b,c$, we get the desired inequality.

$$a^2+b^2+c^2\geq \frac{(a+b+c)^2}{3}$$ thus, it´s sufficient to prove that $$(a+b+c)(a+b+c-6)\geq 27 \Leftrightarrow a+b+c\geq 9$$ and by C-S, $a+b+c=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 9$

Let \[ f(a, b, c) = \sum_{cyc} a^2-2a \ \text{and} \ g(a, b, c) = \sum_{cyc} \frac{1}{a}. \]Using Lagrange Multipliers, we see that $\frac{\partial f}{\partial a} = 2a-2 = \lambda \cdot \frac{\partial g}{\partial a} = \frac{- \lambda}{a^2}$ and similar relation holds for $b$ and $c$. Therefore, $a^3-a^2 = b^3-b^2= c^3-c^2$. Clearly, $\frac{1}{a}, \ \frac{1}{b}, \ \frac{1}{c} < 1$ and thus $a, b, c > 1$. $a^3-a^2=b^3-b^2$ implies $(a-b)(a^2+ab+b^2-a-b)=0$, however, since $a, b > 1$, we have $a^2 + b^2 > a + b \implies a^2+ab+b^2-a-b > 0$. Hence, $a=b$ and similarly $b=c$, implying that the extremum point of a function $f$ is realized for equal values of $a, b$ and $c$. For such values, we have $\frac{1}{a} = \frac{1}{b}=\frac{1}{c} = \frac{1}{3} \implies a=b=c=3$ and $f(3, 3, 3) = 9$. We just need to show that $9$ is minimum. $f(4, 4, 2) = 16 > 9$ and we are done. $\ \square$