By Cauchy-Schwarz inequality: $6(a^2+2b^2+3c^2)=(1+1+1+1+1+1)(a^2+b^2+b^2+c^2+c^2+c^2) \geq (a+2b+3c)^2$ Therefore: $6 \cdot \frac{3}{2} \geq (a+2b+3c)^2 \Leftrightarrow$ $3 \geq a+2b+3c$ Now by AM-GM: $3^{-a}+9^{-b}+27^{-c} \geq 3\sqrt[3]{3^{-a} \cdot 9^{-b} \cdot 27^{-c}}=3\sqrt[3]{3^{-(a+2b+3c)}} \geq 3\sqrt[3]{3^{-3}}=1$

On the bottom line take out $3\sqrt[3]{3^{-a}+9^{-b}+27^{-c}}$ because it is not correct.

Vo Duc Dien wrote: On the bottom line take out $3\sqrt[3]{3^{-a}+9^{-b}+27^{-c}}$ because it is not correct. It's typo. The rest is true.

Now, everything is correct.

If a or b or c are negative it's clearly that $3^{-a}+9^{-b}+27^{-c}\ge1$ So consider $a,b,c \geq 0 $ By Cauchy-Schwartz inequality $\sqrt{(1+2+3)(a^2+2b^2+3c^2)} \geq 1\cdot \sqrt{a^2}+\sqrt{2}\cdot \sqrt{2b^2}+\sqrt{3}\cdot \sqrt{3c^2}$ So, $3 \geq a+2b+3c \geq 0$ By AM-GM $3^{-a}+9^{-b}+27^{-c} \geq 3\sqrt[3]{3^{-(a+2b+3c)}} \geq 3\sqrt[3]{3^{-3}}=1$