cho a,b,c >0, a+2b+3c >20 c/m a+b+c +3/a +9/2b +4/c >13

Giải thích các bước giải:

$\begin{array}{l}
A = a + b + c + \frac{3}{a} + \frac{9}{{2b}} + \frac{4}{c}\\
= \left( {\frac{3}{a} + \frac{{3a}}{4}} \right) + \left( {\frac{9}{{2b}} + \frac{b}{2}} \right) + \left( {\frac{4}{c} + \frac{c}{4}} \right) + \frac{1}{4}\left( {a + 2b + 3c} \right)\\
> 2\sqrt {\frac{3}{a}.\frac{{3a}}{4}} + 2.\sqrt {\frac{9}{{2b}}.\frac{b}{2}} + 2.\sqrt {\frac{4}{c}.\frac{c}{4}} + \frac{1}{4}.20\\
> 3 + 3 + 2 + 5 = 13\\
dau = xay\,ra:\,a = 2;b = 3;c = 4
\end{array}$