Chủ đề này có 5 trả lời

[mathandyou]

Cho $a,b,c>0$ thỏa mãn:$a+b+c=1$.Chứng minh rằng:

$\frac{bc}{\sqrt{a+bc}}+\frac{ca}{\sqrt{b+ca}}+\frac{ab}{\sqrt{c+ab}} \leq \frac{1}{2}$

[Messi10597]

Ta có: $a+bc=a(a+b+c)+bc=(a+b)(a+c)$

$\Rightarrow \frac{bc}{\sqrt{a+bc}}=\frac{bc}{\sqrt{(a+b)(a+c)}}\leq \frac{1}{2}(\frac{bc}{a+b}+\frac{bc}{a+c})$

Tương tự ta có: $\frac{ca}{\sqrt{b+ca}}\leq \frac{1}{2}(\frac{ca}{b+a}+\frac{ca}{b+c})$

                          $\frac{ab}{\sqrt{c+ab}}\leq \frac{1}{2}(\frac{ab}{c+a}+\frac{ab}{c+b})$

Cộng vế với vế , ta có :

                  $\frac{bc}{\sqrt{a+bc}}+\frac{ca}{\sqrt{b+ca}}+\frac{ab}{\sqrt{c+ab}}\leq \frac{1}{2}[\frac{c(a+b)}{a+b}+\frac{a(c+b)}{b+c}+\frac{b(a+c)}{a+c}]=\frac{1}{2}(a+b+c)=\frac{1}{2}$ 

[lymiu]

sao $\frac{1}{2}\left ( a+b+c \right )= \frac{1}{2}$ hở bạn

[hoctrocuanewton]

sao $\frac{1}{2}\left ( a+b+c \right )= \frac{1}{2}$ hở bạn

a+b+c=1 mà bạn 

[lymiu]

ơ ha thanks bạn nha  

[Hoang Tung 126]

Theo AM-GM có :$\sum \frac{bc}{\sqrt{a.1+bc}}=\sum \frac{bc}{\sqrt{a(a+b+c)+bc}}=\sum \frac{bc}{\sqrt{(a+b)(a+c)}}\leq \frac{1}{2}(\sum \frac{bc}{a+b}+\sum \frac{bc}{a+c})=\frac{1}{2}(\sum a)=\frac{1}{2}$