Cho a,b,c>0 thỏa mãn abc=ab+bc+ca
Chứng minh$\frac{1}{a+2b+3c}+\frac{1}{b+2c+3a}+\frac{1}{c+2a+3b}\le \frac{3}{16}$
Cho a,b,c>0 thỏa mãn abc=ab+bc+ca
Chứng minh$\frac{1}{a+2b+3c}+\frac{1}{b+2c+3a}+\frac{1}{c+2a+3b}\le \frac{3}{16}$
Giải thích các bước giải:
Ta có : $abc=ab+bc+ca\rightarrow \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$
Ta có :
$\begin{split}\dfrac{1}{a+2b+3c}&=\dfrac{1}{(a+2b)+3c}\\&\le \dfrac{1}{4}(\dfrac{1}{a+2b}+\dfrac{1}{3c})\\&=\dfrac{1}{4}(\dfrac{1}{a+b+b}+\dfrac{1}{3c})\\ &=\\&\le \dfrac{1}{4}.(\dfrac{1}{9}.(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b})+\dfrac{1}{3c})\\&=\dfrac{1}{36a}+\dfrac{1}{18b}+\dfrac{1}{12c}\end{split}$
Chứng minh tương tự ta có :
$\begin{cases}\dfrac{1}{b+2c+3a}\le \dfrac{1}{36b}+\dfrac{1}{18c}+\dfrac{1}{12a}\\\dfrac{1}{c+2a+3b}\le \dfrac{1}{36c}+\dfrac{1}{18a}+\dfrac{1}{12b}\end{cases}$
$\rightarrow \dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{1}{36a}+\dfrac{1}{18b}+\dfrac{1}{12c}+ \dfrac{1}{36b}+\dfrac{1}{18c}+\dfrac{1}{12a}+\dfrac{1}{36c}+\dfrac{1}{18a}+\dfrac{1}{12b}$
$\rightarrow \dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{1}{6}(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})=\dfrac{1}{6}<\dfrac{3}{16}$