A=a2a+2b2+b2b+2c2+c2c+2a2Ta có: a2a+2b2+(a+2b2)≥2a (BĐT Cô-si)b2b+2c2+(b+2c2)≥2bc2c+2a2+(c+2a2)≥2c⇒A+a+b+c+2(a2+b2+c2)≥2(a+b+c)⇒A≥3−2(a2+b2+c2)Theo BĐT Bunhia: a+b+c≤3(a2+b2+c2)⇒a2+b2+c2≥1⇒A≥1(đpcm)
A=a2a+2b2+b2b+2c2+c2c+2a2Ta có: a2a+2b2+(a+2b2)≥2a (BĐT Cô-si)b2b+2c2+(b+2c2)≥2bc2c+2a2+(c+2a2)≥2c⇒A+a+b+c+2(a2+b2+c2)≥2(a+b+c)⇒A≥3−2(a2+b2+c2)Theo BĐT Bunhia: $(a+b+c)^2\leq3(a^2+b^2+c^2)⇒a2+b2+c2≥1\Rightarrow A \ge 1 (đpcm)$