bạn làm rõ đoạn tô đỏ cái
$\frac{1}{a^{2}+b^{2}+c^{2}}+\frac{1}{abc}= \frac{1}{a^{2}+b^{2}+c^{2}}+\frac{1}{3abc}+\frac{2}{3abc}= \frac{1}{a^{2}+b^{2}+c^{2}}+ \frac{(a+b+c)^{2}}{3abc}+ \frac{2a+2b+2c}{3abc}= \frac{1}{a^{2}+b^{2}+c^{2}}+ \frac{a^{2}+b^{2}+c^{2}}{3abc}+ \frac{2ab+2bc+2ca}{3abc}+ \frac{2}{abc}= \frac{1}{a^{2}+b^{2}+c^{2}}+ \frac{a^{2}+b^{2}+c^{2}}{3abc}+ \frac{2}{3a}+ \frac{2}{3b}+\frac{2}{3c}+ \frac{2}{3abc}\geq \frac{2}{\sqrt{3abc}}+ \frac{18}{3a+3b+3c}+ \frac{18}{(a+b+c)^{3}}\geq 6+6+18=30$
đủ rõ chưa
- hoangtubatu955 và leduylinh1998 thích