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Ta có : \(T\ge2\sqrt{\sqrt{\left(x^2-x+2\right)\left(x^2+x+2\right)}}=2\sqrt[4]{\left(x^2-x+2\right)\left(x^2+x+2\right)}\)

\(=2\sqrt[4]{x^4+3x^2+4}\ge2\sqrt[4]{4}=2\sqrt{2}\)

Vậy Min T = \(2\sqrt{2}\)khi x = 0

\(P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ba}+\frac{c^2}{ac+bc}\)

\(\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)

\(\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)

dấu "=" xảy ra tại a=b=c

Cách 2

\(P+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)

\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)

\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\left(a+b+c\right)\cdot\frac{9}{2\left(a+b+c\right)}=\frac{9}{2}\)

\(\Rightarrow P\ge\frac{3}{2}\Leftrightarrow a=b=c\)

\(P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\left(1\right)\)

Đặt \(\hept{\begin{cases}b+c=x\\c+a=y\\a+b=z\end{cases}\left(x,y,z>0\right)}\)

\(\Rightarrow a=\frac{y+z-x}{2}\);\(b=\frac{z+x-y}{2}\);\(c=\frac{x+y-z}{2}\)

\(\left(1\right)\)trở thành \(\frac{y+z-x}{2x}+\frac{z+x-y}{2y}+\frac{x+y-z}{2z}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{y}{2x}+\frac{z}{2x}-\frac{1}{2}+\frac{z}{2y}+\frac{x}{2y}-\frac{1}{2}+\frac{x}{2z}+\frac{y}{2z}-\frac{1}{2}\ge\frac{3}{2}\)

\(\Leftrightarrow\left(\frac{y}{2x}+\frac{x}{2y}\right)+\left(\frac{z}{2x}+\frac{x}{2z}\right)+\left(\frac{z}{2y}+\frac{y}{2z}\right)\ge3\)

Vì \(\frac{y}{2x}+\frac{x}{2y}\ge2\sqrt{\frac{y}{2x}.\frac{x}{2y}}=1\)( bđt AM-GM)

CMTT ​​\(\frac{z}{2x}+\frac{x}{2z}\ge1\)và \(\frac{z}{2y}+\frac{y}{2z}\ge1\)

rồi cộng vào là xong

Dấu"="xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{y}{2x}=\frac{x}{2y}\\\frac{z}{2x}=\frac{x}{2z}\\\frac{z}{2y}=\frac{y}{2z}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}2x^2=2y^2\\2z^2=2x^2\\2y^2=2z^2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=y\\z=x\\y=z\end{cases}\Leftrightarrow}x=y=z\)

Vậy \(P_{min}=\frac{3}{2}\Leftrightarrow x=y=z\)

*Tìm min:

Ta có: \(F-\left(xy+yz+zx\right)=\frac{3}{4}\left(x-y\right)^2+\frac{1}{4}\left(x+y-2z\right)^2\ge0\)

Do đó: \(F\ge xy+yz+zx=11\)

Đẳng thức xảy ra khi \(x=y=z=\sqrt{\frac{11}{3}}\)

*Tìm max: Chưa nghĩ ra.