Bài 2:
a) CM $\sum a^2-\sum ab\geq 3(a-b)(b-c)$
Ta thấy $\text{VT}=\frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$
Dễ thấy $(a-b)^2+(b-c)^2-2(a-b)(b-c)=(a-b+b-c)^2\geq 0\Rightarrow (a-b)^2+(b-c)^2\geq 2(a-b)(b-c)$
Mặt khác $(c-a)^2=(a-c)^2=(a-b+b-c)^2\geq 4(a-b)(b-c)$
$\Rightarrow \text{VT}\geq 3(a-b)(b-c)$
b) CM $\sum \sqrt{\frac{ab+2c^2}{1+ab-c^2}}\geq 2+ab+bc+ac$
Áp dụng BĐT AM-GM:
$\text{VT}=\sum \sqrt{\frac{ab+2c^2}{a^2+ab+b^2}}=\sum \frac{ab+2c^2}{\sqrt{(a^2+ab+b^2)(ab+2c^2)}}$
$\geq \sum \frac{ab+2c^2}{\sqrt{\left ( \frac{a^2+b^2+2ab+2c^2}{2} \right )^2}}=\sum \frac{2(ab+2c^2)}{(a+b)^2+2c^2}\geq \sum \frac{2(ab+2c^2)}{2(a^2+b^2+c^2)}=\text{VP}$
Ta có đpcm
Bài 3:
BĐT cần CM $\Leftrightarrow [\sum c^2(a^2+b^2)^2](a+b+c)^4\sum (ab)^4\geq 54^2(abc)^6$
Thuần AM-GM:
$\text{VT}\geq (\sum 4c^2a^2b^2)[3\sqrt[3]{abc}]^4[3\sqrt[3]{a^8b^8c^8}]=12(abc)^2.3^5\sqrt[3]{a^{12}b^{12}c^{12}}=54^2(abc)^6=\text{VP}$