Ta có $(a^3-b^3)(a^2-b^2)\geq 0\Rightarrow a^5+b^5\geq a^2b^2(a+b)$
$\Rightarrow \frac{ab}{a^5+b^5+ab}\leq \frac{ab}{a^2b^2(a+b)+ab}\times \frac{c^2}{c^2}\leq \frac{c}{a+b+c}$ ( vì abc=1)
Tương tự :$\frac{bc}{b^5+c^5+bc}\leq \frac{bc}{b^2c^2(b+c)+bc}\times \frac{a^2}{a^2}\leq \frac{a}{a+b+c}$
$\frac{ac}{a^5+c^5+ca}\leq \frac{ca}{c^2a^2(a+c)+ac}\times \frac{c^2}{c^2}\leq \frac{b}{a+b+c}$
Cộng 3 BĐT trên lại theo từng vế ta => ĐPCM
Dấu "=" xảy ra khi a=b=c=1