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