CM: Với $0\leq$$a$$\leq$$b$$\leq$$c$ thì $\frac{a^{2005}+b^{2005}+c^{2005}}{a^{2006}+b^{2006}+c^{2006}}$$\leq $$\frac{3}{a+b+c}$
Ta có :$VT=\frac{a^{2005}+b^{2005}+c^{2005}}{a^{2006}+b^{2006}+c^{2006}} \le \frac{a^{2004}+b^{2004}+c^{2004}}{a^{2005}+b^{2005}+c^{2005}} \Leftrightarrow (\sum a^{2005})^2 \le \sum a^{2004}.\sum a^{2006}$
(đúng theo bđt cauchy-schwarz)
cmtt $ \frac{a^{2004}+b^{2004}+c^{2004}}{a^{2005}+b^{2005}+c^{2005} } \le \frac{a^{2003}+b^{2003}+c^{2003}}{a^{2004}+b^{2004}+c^{2004}}$
.........
Từ các bđt trên ta suy ra $VT \le  \frac{a^0+b^0+c^0}{a^1+b^1+c^1}=VP$
Vậy ta có đpcm và đẳng thức xảy ra $\Leftrightarrow a=b=c$
chèn ơi,ra vậy ....@@@ –  Ngọc 14-04-16 10:27 PM
vậy đó :D –  tran85295 14-04-16 10:27 PM
$\sum a^{2005} =a^{2005} b^{2005} c^{2005}$ –  tran85295 14-04-16 10:26 PM
viết tắt ấy mà –  tran85295 14-04-16 10:26 PM
còn cách khác ko, chỉ e vs –  Ngọc 14-04-16 10:25 PM
mấy cái kí hiệu j z? em ko hiểu –  Ngọc 14-04-16 10:25 PM
Đặt $x=a^{2015};y=b^{2015};z=c^{2015}$ khi đó BPT thành:
$\frac{x+y+z}{ax+by+cz}\leq \frac{3}{a+b+c}\Rightarrow a(y+z-2x)+b(x+z-2y)+c(x+y-2z)\leq 0$
$\Rightarrow a(y-x)+a(z-x)+b(x-y)+b(z-y)+c(x-z)+c(y-z)\leq 0$
$\Rightarrow (y-x)(a-b)+(z-x)(a-c)+(x-y)(b-c)\leq 0$(1)
Có $(y-x)(a-b)=(b^{2015}-a^{2015})(a-b)=-(a-b)^{2}(a^{2014}+a^{2013}b+...+b^{2014})\leq 0$
CMTT $\Rightarrow $(1)$\Rightarrow $ ĐPCM

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