Chủ đề này có 1 trả lời

[luukhaiuy]

cho a,b,c>0 CMR $\frac{ab}{4b+4c+a}+\frac{bc}{4c+4a+b}+\frac{ca}{4a+4b+c}\leq \frac{a+b+c}{9}$

[leminhnghiatt]

cho a,b,c>0 CMR $\frac{ab}{4b+4c+a}+\frac{bc}{4c+4a+b}+\frac{ca}{4a+4b+c}\leq \frac{a+b+c}{9}$

Ta có:

 

$\dfrac{ab}{4b+4c+a}=\dfrac{ab}{(2b+c)+(2b+c)+(2c+a)}=\dfrac{ab}{9}.\dfrac{9}{(2b+c)+(2b+c)+(2c+a)}$

 

Áp dụng bđt: $\dfrac{9}{x+y+z} \leq \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$

 

Ta có: $\dfrac{ab}{9}.\dfrac{9}{(2b+c)+(2b+c)+(2c+a)} \leq \dfrac{ab}{9}[\dfrac{1}{2b+c}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}]=\dfrac{ab}{9}[\dfrac{2}{2b+c}+\dfrac{1}{2c+a}]$

 

$ \longrightarrow \dfrac{ab}{4b+4c+a} \leq \dfrac{ab}{9}[\dfrac{2}{2b+c}+\dfrac{1}{2c+a}]$ (1)

 

Chứng minh TT ta có: $\dfrac{bc}{4c+4a+b} \leq \dfrac{bc}{9}[\dfrac{2}{2c+a}+\dfrac{1}{2a+b}]$ (2)

 

$\dfrac{ca}{4a+4b+c} \leq \dfrac{ca}{9}[\dfrac{2}{2a+b}+\dfrac{1}{2b+c}]$ (3)

 

Cộng các bất đẳng thức lại ta có:

 

$VT \leq \dfrac{2ab+ca}{9(2b+c)}+\dfrac{2bc+ab}{9(2c+a)}+\dfrac{2ca+bc}{9(2a+b)}$

 

$\iff VT \leq \dfrac{a}{9}+\dfrac{b}{9}+\dfrac{c}{9}$

 

$\iff VT \leq \dfrac{a+b+c}{9}$

 

Dấu "=" xảy ra khi: $a=b=c$