Proving $(a+b+c) \Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big) \leqslant 25$

By AM-GM we have $$ \frac{(a+b+c) + (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}{2} \geq \sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}. $$ Note that by the assumption, we have $$ 3 + \frac{1}{3} \geq a + \frac{1}{a} $$ and similarly for the other variables. Therefore $$ 3 \cdot \frac{10}{3} \cdot \frac{1}{2} \geq \sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}, $$ as desired.

I found a better estimation $$ (a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant \frac{209}{9}.$$ Equality occur when $a=b=3,\,c=\frac 13$ or $a=b=\frac 13,\,c=3.$