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[SOYA264]

Cho $a,b,c$ là 3 số thực dương. Chứng minh rằng:

 

$\frac{a^2}{5a^2+(b+c)^2}+\frac{b^2}{5b^2+(c+a)^2}+\frac{c^2}{5c^2+(a+b)^2}\leq \frac{1}{3}$

 

 

Edited by SOYA264, 24-09-2013 - 12:59.

[Hoang Tung 126]

Ta có :$\sum \frac{9a^2}{5a^2+(b+c)^2}=\sum \frac{(a+a+a)^2}{(a^2+b^2+c^2)+(2a^2+bc)+(2a^2+bc)}\leq \sum \frac{a^2}{a^2+b^2+c^2}+\sum \frac{a^2}{2a^2+bc}+\sum \frac{a^2}{2a^2+bc}=\sum \frac{a^2}{a^2+b^2+c^2}+2.\sum \frac{a^2}{2a^2+bc}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}+2.(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab})$(1) .

Mặt khác theo bđt Bunhiacopxki ta có :$\sum \frac{bc}{2a^2+bc}=\sum \frac{b^2c^2}{b^2c^2+2a^2bc}\geq \frac{(ab+bc+ac)^2}{(a^2b^2+b^2c^2+c^2a^2+2abc.(a+b+c))}=\frac{(ab+bc+ac)^2}{(ab+bc+ac)^2}=1$ nên $\sum (1-\frac{bc}{2a^2+bc})\leq 2$ hay $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\leq 1$(2) .Từ (1) và (2) $= > \sum \frac{9a^2}{5a^2+b^2+c^2}\leq 1+2.(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab})\leq 1+2.1=3$ nên $\frac{a^2}{5a^2+(b+c)^2}+\frac{b^2}{5b^2+(a+c)^2}+\frac{c^2}{5c^2+(a+b)^2}\leq \frac{3}{9}=\frac{1}{3}$(đpcm) .

 Dấu = xảy ra khi a=b=c

Edited by Hoang Tung 126, 24-09-2013 - 15:19.