Ta có
$\frac{a^5-a^2}{a^5+b^2+c^2}=\frac{a^5+b^2+c^2-(a^2+b^2+c^2)}{a^5+b^2+c^2}=1-\frac{a^2+b^2+c^2}{a^5+b^2+c^2}$
BĐT TƯƠNG ĐƯƠNG
$3-\left ( a^2+b^2+c^2 \right )\left ( \frac{1}{a^5+b^2+c^2}+\frac{1}{b^5+c^2+a^2}+\frac{1}{c^5+a^2+b^2} \right )\geq 0$
Theo BĐT BCS
$(a^5+b^2+c^2)(\frac{1}{a}+b^2+c^2)\geq (a^2+b^2+c^2)^2\Rightarrow a^5+b^2+c^2\geq \frac{(a^2+b^2+c^2)^2}{\frac{1}{a}+b^2+c^2}$
SUY RA
VT$\geq 3-\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2(a^2+b^2+c^2)}{a^2+b^2+c^2}=1-\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{a^2+b^2+c^2}=1-\frac{ab+bc+ca}{abc(a^2+b^2+c^2)}\geq 1-\frac{ab+bc+ca}{abc(ab+bc+ca)}=1-1=0$