2 replies to this topic

[leduylinh1998]

Cho a,b,c là các số dương thỏa mãn:$a^{2}+b^{2}+c^{2}=1$. Tìm Max:

$M=\sum \frac{1}{1-bc}$

[kfcchicken98]

ta có $M=\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}$

$\frac{1}{1-ab}-1=\frac{ab}{1-ab}$

suy ra M-3=$\frac{ab}{1-ab}+\frac{bc}{1-bc}+\frac{ca}{1-ac}$

ta có $2M-6=\frac{4ab}{2-2ab}+\frac{4bc}{2-2bc}+\frac{4ca}{2-2ca}=\frac{4ab}{(a-b)^{2}+c^{2}+1}+\frac{4bc}{(b-c)^{2}+1+a^{2}}+\frac{4ca}{(c-a)^{2}+1+b^{2}}\leq \frac{(a+b)^{2}}{a^{2}+c^{2}+b^{2}+c^{2}}+\frac{(b+c)^{2}}{a^{2}+b^{2}+a^{2}+c^{2}}+\frac{(c+a)^{2}}{c^{2}+b^{2}+a^{2}+b^{2}}\leq \frac{a^{2}}{a^{2}+c^{2}}+\frac{b^{2}}{b^{2}+c^{2}}+\frac{b^{2}}{a^{2}+b^{2}}+\frac{c^{2}}{a^{2}+c^{2}}+\frac{c^{2}}{b^{2}+c^{2}}+\frac{a^{2}}{a^{2}+b^{2}}=3$

suy ra $2M-6\leq 3$

suy ra $M\leq \frac{9}{2}$

[hoctrocuanewton]

ta có

$\sum \frac{1}{1-bc}=3+\sum \frac{bc}{1-bc}$ (1)

ta lại có

$\sum \frac{bc}{1-bc}\leq\sum \frac{2bc}{2-b^{2}-c^{2}}= \sum \frac{4bc}{2(2a^{2}+b^{2}+c^{2})}\leq \frac{(b+c)^{2}}{2(2a^{2}+b^{2}+c^{2})}$ (2)

 

áp dụng bđt schwars ta có

$\sum \frac{(b+c)^{2}}{2(2a^{2}+b^{2}+c^{2})}\leq \frac{b^{2}}{2(a^{2}+b^{2})}+\frac{c^{2}}{2(a^{2}+c^{2})}+\frac{a^{2}}{2(a^{2}+b^{2})}+\frac{c^{2}}{2(b^{2}+c^{2})}+\frac{b^{2}}{2(b^{2}+c^{2})}+\frac{a^{2}}{2(a^{2}+c^{2})}$

$\Rightarrow \sum \frac{(b+c)^{2}}{2(2a^{2}+b^{2}+c^{2})}\leq \frac{3}{2}$ (3)

(1)(2) suy ra Max của M=$\frac{9}{2}$

dấu bằng xảy ra khi $a=b=c=\frac{\sqrt{3}}{3}$

Edited by hoctrocuanewton, 24-12-2013 - 12:31.