Using the problem's condition, multiply the given inequality by $2$ and add $a^2+3b^2+5c^2$ to each side to obtain $$\left(\frac{1}{a}+\frac{1}{a}+a^2\right)+3\left(\frac{1}{b}+\frac{1}{b}+b^2\right)+5\left(\frac{1}{c}+\frac{1}{c}+c^2\right)\geq 9(a^2+b^2+c^2)=27$$Now using the AM-GM inequality we obtain $\frac{1}{x}+\frac{1}{x}+x^2\geq3\sqrt[3]{\frac{1}{x}\cdot\frac{1}{x}\cdot x^2}=3$ for all $x>0$. Applying this for $x=a, x=b, x=c$, we have that $$LHS\geq 3+3\cdot 3+5\cdot 3=27$$which finishes our proof. Equality is reached if and only if $a=b=c=1$.