How can I prove that $(\frac1a+\frac1b+\frac1c)(\sqrt a+\sqrt b+\sqrt c)^2\ge 3^3$ using Hölder's Inequality
This cannot be done with the usual Hölder inequality, there is a negative exponent and $2$ between $0$ and $1$. And I couldn't find any generalization with those exponents. (There's one formula which holds for a different case here)
Hölder is the following.
(In the following form much easier to use Hölder!)
Let $a_i>0$, $b_i>0$, $\alpha>0$ and $\beta>0$. Hence, we have:
$$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq$$ $$\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$$
For your inequality $n=3$, $\alpha=1$ and $\beta=2$ and it's just Hölder!