$\frac{x^2}{y^2}+1\ge2|\frac xy|\geq 2\frac{x}{y}$tuong tu cong lai$VT\ge 2(\frac xy+\frac yz+\frac zx)-3\ge VP$ (AM-GM 3 so)

$\frac{x^2}{y^2}+1\ge2|\frac xy|$tuong tu cong lai$VT\ge 2(|\frac xy|+|\frac yz|+|\frac zx|)-3\ge \frac xy + \frac yz + \frac zx+(|\frac xy|+|\frac yz|+|\frac zx|-3)\ge VP$ (AM-GM 3 so)

$M=3+\frac{a}b+\frac{b}a+\frac{b}c+\frac{c}b+\frac{c}a+\frac{a}c$Ta có $\frac14\le\frac{a}c<4\Rightarrow (\frac14-\frac{a}c)(4-\frac{a}c)\le0\Leftrightarrow 1+\frac{a^2}{c^2}\le\frac{17}{4}.\frac{a}c\Leftrightarrow \frac{a}c+\frac{c}a\le\frac{17}{4}$Ta cũng có $(1-\frac{a}b)(1-\frac{b}c)+(1-\frac{b}a)(1-\frac{c}b)\ge0$ ( do mỗi số hạng $\ge0$)$\Leftrightarrow\frac{a}b+\frac{b}a+\frac{b}c+\frac{c}b \le 2+\frac{a}c+\frac{c}a=2+\frac{17}4=\frac{25}4$cộng lại ta có $M\le3+\frac{17}4+\frac{25}4=\frac{27}4$$\Leftrightarrow (a;b;c)=(1;1;4)$ hoặc $(1;4;4)$

$P=3+\frac{a}b+\frac{b}a+\frac{b}c+\frac{c}b+\frac{c}a+\frac{a}c$Ta có $\frac14\le\frac{a}c<4\Rightarrow (\frac14-\frac{a}c)(4-\frac{a}c)\le0\Leftrightarrow 1+\frac{a^2}{c^2}\le\frac{17}{4}.\frac{a}c\Leftrightarrow \frac{a}c+\frac{c}a\le\frac{17}{4}$Ta cũng có $(1-\frac{a}b)(1-\frac{b}c)+(1-\frac{b}a)(1-\frac{c}b)\ge0$ ( do mỗi số hạng $\ge0$)$\Leftrightarrow\frac{a}b+\frac{b}a+\frac{b}c+\frac{c}b \le 2+\frac{a}c+\frac{c}a=2+\frac{17}4=\frac{25}4$cộng lại ta có $P\le3+\frac{17}4+\frac{25}4=\frac{27}4$Dấu bằng xảy ra $\Leftrightarrow (a;b;c)=(1;1;4)$ hoặc $(1;4;4)$