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Giải:
Ta có: 1 = $\dfrac{1}{a}$ + $\dfrac{2}{b}$ + $\dfrac{3}{c}$ \geq $\dfrac{(1 + \sqrt[]{2} + \sqrt[]{3})^2}{a+b+c}$

\Rightarrow a + b + c \geq $(1 + \sqrt[]{2} + \sqrt[]{3})^2$

Dấu "=" xảy ra khi chỉ khi: $\dfrac{1}{a}$ = $\dfrac{\sqrt[]{2}}{b}$ = $\dfrac{\sqrt[]{3}}{c}$ và $\dfrac{1}{a}$ + $\dfrac{2}{b}$ + $\dfrac{3}{c}$ = 1

\Leftrightarrow (a;b;c) = $( \sqrt[]{3}+ \sqrt[]{2}+1; \sqrt[]{2}+ \sqrt[]{6}+2; \sqrt[]{3}+ \sqrt[]{6}+3)$
 
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