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transformers123

Ta có

$A= a+b+c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}$

$\iff A= \dfrac{3a}{4}+\dfrac{3}{a}+\dfrac{c}{4}+\dfrac{4}{c}+\dfrac{b}{2}+\dfrac{9}{2b}+\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}$

$\iff A \ge 2\sqrt{\dfrac{9a}{4a}}+2\sqrt{\dfrac{4c}{4c}}+2 \sqrt{\dfrac{9b}{4b}}+\dfrac{a+2b+3c}{4}$

$\iff A \ge 2.\dfrac{3}{2}+2+2.\dfrac{3}{2}+5$

$\iff A \ge 13$

Dấu "=" xảy ra khi $\begin{cases}a=2\\b=3\\c=4\end{cases}$
 
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